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LSpice
LSpice
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This is not precisely an answer to your question (since you would like yoto understand Kneser's paper), but at the expense of using some modern technology (mainly sheaf cohomology) and being a little less constructive, I think one can get a very short proof of the result you mention.

EDIT: as abx points out in a comment below, my proof does not quite work. A fourth hypersurface containing $C_0$ of sufficiently high degree is needed to get rid of the potential points of intersection of $S_3$ with the $C_i$ ($i>0$) which do not lie on $C_0$.

Let $\mathcal{C} \subset \mathbb{P}^{3}$ be an integral projective curve (over an algebraically closed field) and denote by $\mathcal{I}$ its sheaf of ideal in $\mathbb{P}^3$. For any $d \geq 0$, the vector space:

$$H^0(\mathbb{P}^3, \mathcal{I}(d))$$

is the vector space of hypersurfaces of degree $d$ which contain $\mathcal{C}$. By Serre's vanishing Theorem, we know that the sheaf $\mathcal{I}(d_1)$ is globally generated for $d_1>>0$$d_1\gg0$, in particular $H^0(\mathbb{P}^3, \mathcal{I}(d_1)) \neq 0$ for $d_1>>0$$d_1\gg0$. An explicit bound for $d_1$ should be equivalent to the computation of the degree of your $f$ generating $I \cap k[x,y,z]$.

Let $f_1, f_2$ be two generic elements of $H^0(\mathbb{P}^3, \mathcal{I}(d_1))$. Up to increasing $d_1$, we can assume that the hypersurfaces $S_1$ and $S_2$ given by $f_1 = 0$ and $f_2 =0$ are integral. So in particular, we have, set-theoretically:

$$ S_1 \cap S_2 = \bigcup_{i=0}^{r} \mathcal{C}_i,$$ with $\mathcal{C}_0 = \mathcal{C}$ and the $\mathcal{C}_i$ are integral projective curves that meet properly.

Now, it is sufficient to find $d_2>0$ such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2))$ does not contain any of the $\mathcal{C}_i$ for $i>0$. Since $k$ is infinite and any vector space over $k$ is thus infinte, it is sufficient to prove that given a fixed $i \neq 0$, there exists $d_2(i) >0$, such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2(i)))$ does not contain $\mathcal{C}_i$ (and then apply the result for all $i$ by taking the complement in $H^0(\mathbb{P}^3, \mathcal{I}(d_3))$ of the union of the subspaces to be avoided, where $d_3$ is the max of the $d_2(i)$).

Thus, we are left to prove that for $d_2>>0$$d_2\gg0$, we have: $$ H^0(\mathbb{P}^3, \mathcal{I}(d_2)) \neq H^0(\mathbb{P}^3, \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)),$$ where $\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}$ is the ideal sheaf of $\mathcal{C}_0 \cup \mathcal{C}_i$. Since the intersection $\mathcal{C}_0 \cap \mathcal{C}_i$ has the right codimension (i.e. $3$), we have an exact sequence (see Sasha's answer to the first answer to this questionProducts of Ideal Sheaves and Union of irreducible Subvarieties):

$$0 \longrightarrow \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}} \longrightarrow \mathcal{O}_{\mathbb{P}^3} \longrightarrow \mathcal{O}_{\mathcal{C}_0} \oplus \mathcal{O}_{\mathcal{C}_i} \longrightarrow \mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i} \longrightarrow 0.$$$$0 \longrightarrow \mathcal{I}_{\mathcal{C}_0 \cup \mathcal{C}_i} \longrightarrow \mathcal{O}_{\mathbb{P}^3} \longrightarrow \mathcal{O}_{\mathcal{C}_0} \oplus \mathcal{O}_{\mathcal{C}_i} \longrightarrow \mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i} \longrightarrow 0.$$

Tensoring by $\mathcal{O}_{\mathbb{P}^3}(d_2)$ and taking Euler carateristics, we get: $$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{O}_{\mathbb{P}^3}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_i}(d_2)) + \chi(\mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ that is:

$$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ where $\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}$ is the ideal sheaf of $\mathcal{C}_0 \cap \mathcal{C}_i$ in $\mathcal{O}_{\mathcal{C}_i}$.

Applying Serre's vanishing theorem one more time, we find that for $d_2>>0$$d_2\gg0$:

$\bullet$ $\chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = H^0(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2))$, $\chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) = H^0(\mathcal{J}_{\mathcal{C}_0}(d_2))$ and $\chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) = H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2))$,

$\bullet$ $H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) \neq 0$.

  • $\chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = H^0(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2))$, $\chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) = H^0(\mathcal{J}_{\mathcal{C}_0}(d_2))$ and $\chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) = H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2))$,

  • $H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) \neq 0$.

As a consequence, for $d_2(i)>>0$$d_2(i)\gg0$, the generic hypersurface of degree $d_2(i)$ containing $\mathcal{C}_0$ does not contain $\mathcal{C}_i$. This is true for all $i$ and since $k$ is infinite, by taking $d_3$ the max of the $d_2(i)$, we find that the generic hypersurface of degree $d_3$ does not contain any of the $\mathcal{C}_i$ for $i \neq 0$. Let $S_3$ be such a generic hypersurface, then we have, set-theoretically:

$$S_1 \cap S_2 \cap S_3 = \mathcal{C}_0.$$

EDIT : as abx points out in the comment below, my proof does not quite work. A fourth hypersurface containing $C_0$ of sufficiently high degree is needed to get rid off the potentiel points of intersection of $S_3$ with the $C_i$ ($i>0$) which do not lie on $C_0$.

This is not precisely an answer to your question (since you would like yo understand Kneser's paper), but at the expense of using some modern technology (mainly sheaf cohomology) and being a little less constructive, I think one can get a very short proof of the result you mention.

Let $\mathcal{C} \subset \mathbb{P}^{3}$ be an integral projective curve (over an algebraically closed field) and denote by $\mathcal{I}$ its sheaf of ideal in $\mathbb{P}^3$. For any $d \geq 0$, the vector space:

$$H^0(\mathbb{P}^3, \mathcal{I}(d))$$

is the vector space of hypersurfaces of degree $d$ which contain $\mathcal{C}$. By Serre's vanishing Theorem, we know that the sheaf $\mathcal{I}(d_1)$ is globally generated for $d_1>>0$, in particular $H^0(\mathbb{P}^3, \mathcal{I}(d_1)) \neq 0$ for $d_1>>0$. An explicit bound for $d_1$ should be equivalent to the computation of the degree of your $f$ generating $I \cap k[x,y,z]$.

Let $f_1, f_2$ be two generic elements of $H^0(\mathbb{P}^3, \mathcal{I}(d_1))$. Up to increasing $d_1$, we can assume that the hypersurfaces $S_1$ and $S_2$ given by $f_1 = 0$ and $f_2 =0$ are integral. So in particular, we have, set-theoretically:

$$ S_1 \cap S_2 = \bigcup_{i=0}^{r} \mathcal{C}_i,$$ with $\mathcal{C}_0 = \mathcal{C}$ and the $\mathcal{C}_i$ are integral projective curves that meet properly.

Now, it is sufficient to find $d_2>0$ such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2))$ does not contain any of the $\mathcal{C}_i$ for $i>0$. Since $k$ is infinite and any vector space over $k$ is thus infinte, it is sufficient to prove that given a fixed $i \neq 0$, there exists $d_2(i) >0$, such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2(i)))$ does not contain $\mathcal{C}_i$ (and then apply the result for all $i$ by taking the complement in $H^0(\mathbb{P}^3, \mathcal{I}(d_3))$ of the union of the subspaces to be avoided, where $d_3$ is the max of the $d_2(i)$).

Thus, we are left to prove that for $d_2>>0$, we have: $$ H^0(\mathbb{P}^3, \mathcal{I}(d_2)) \neq H^0(\mathbb{P}^3, \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)),$$ where $\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}$ is the ideal sheaf of $\mathcal{C}_0 \cup \mathcal{C}_i$. Since the intersection $\mathcal{C}_0 \cap \mathcal{C}_i$ has the right codimension (i.e. $3$), we have an exact sequence (see the first answer to this question):

$$0 \longrightarrow \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}} \longrightarrow \mathcal{O}_{\mathbb{P}^3} \longrightarrow \mathcal{O}_{\mathcal{C}_0} \oplus \mathcal{O}_{\mathcal{C}_i} \longrightarrow \mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i} \longrightarrow 0.$$

Tensoring by $\mathcal{O}_{\mathbb{P}^3}(d_2)$ and taking Euler carateristics, we get: $$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{O}_{\mathbb{P}^3}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_i}(d_2)) + \chi(\mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ that is:

$$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ where $\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}$ is the ideal sheaf of $\mathcal{C}_0 \cap \mathcal{C}_i$ in $\mathcal{O}_{\mathcal{C}_i}$.

Applying Serre's vanishing theorem one more time, we find that for $d_2>>0$:

$\bullet$ $\chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = H^0(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2))$, $\chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) = H^0(\mathcal{J}_{\mathcal{C}_0}(d_2))$ and $\chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) = H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2))$,

$\bullet$ $H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) \neq 0$.

As a consequence, for $d_2(i)>>0$, the generic hypersurface of degree $d_2(i)$ containing $\mathcal{C}_0$ does not contain $\mathcal{C}_i$. This is true for all $i$ and since $k$ is infinite, by taking $d_3$ the max of the $d_2(i)$, we find that the generic hypersurface of degree $d_3$ does not contain any of the $\mathcal{C}_i$ for $i \neq 0$. Let $S_3$ be such a generic hypersurface, then we have, set-theoretically:

$$S_1 \cap S_2 \cap S_3 = \mathcal{C}_0.$$

EDIT : as abx points out in the comment below, my proof does not quite work. A fourth hypersurface containing $C_0$ of sufficiently high degree is needed to get rid off the potentiel points of intersection of $S_3$ with the $C_i$ ($i>0$) which do not lie on $C_0$.

This is not precisely an answer to your question (since you would like to understand Kneser's paper), but at the expense of using some modern technology (mainly sheaf cohomology) and being a little less constructive, I think one can get a very short proof of the result you mention.

EDIT: as abx points out in a comment below, my proof does not quite work. A fourth hypersurface containing $C_0$ of sufficiently high degree is needed to get rid of the potential points of intersection of $S_3$ with the $C_i$ ($i>0$) which do not lie on $C_0$.

Let $\mathcal{C} \subset \mathbb{P}^{3}$ be an integral projective curve (over an algebraically closed field) and denote by $\mathcal{I}$ its sheaf of ideal in $\mathbb{P}^3$. For any $d \geq 0$, the vector space:

$$H^0(\mathbb{P}^3, \mathcal{I}(d))$$

is the vector space of hypersurfaces of degree $d$ which contain $\mathcal{C}$. By Serre's vanishing Theorem, we know that the sheaf $\mathcal{I}(d_1)$ is globally generated for $d_1\gg0$, in particular $H^0(\mathbb{P}^3, \mathcal{I}(d_1)) \neq 0$ for $d_1\gg0$. An explicit bound for $d_1$ should be equivalent to the computation of the degree of your $f$ generating $I \cap k[x,y,z]$.

Let $f_1, f_2$ be two generic elements of $H^0(\mathbb{P}^3, \mathcal{I}(d_1))$. Up to increasing $d_1$, we can assume that the hypersurfaces $S_1$ and $S_2$ given by $f_1 = 0$ and $f_2 =0$ are integral. So in particular, we have, set-theoretically:

$$ S_1 \cap S_2 = \bigcup_{i=0}^{r} \mathcal{C}_i,$$ with $\mathcal{C}_0 = \mathcal{C}$ and the $\mathcal{C}_i$ are integral projective curves that meet properly.

Now, it is sufficient to find $d_2>0$ such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2))$ does not contain any of the $\mathcal{C}_i$ for $i>0$. Since $k$ is infinite and any vector space over $k$ is thus infinte, it is sufficient to prove that given a fixed $i \neq 0$, there exists $d_2(i) >0$, such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2(i)))$ does not contain $\mathcal{C}_i$ (and then apply the result for all $i$ by taking the complement in $H^0(\mathbb{P}^3, \mathcal{I}(d_3))$ of the union of the subspaces to be avoided, where $d_3$ is the max of the $d_2(i)$).

Thus, we are left to prove that for $d_2\gg0$, we have: $$ H^0(\mathbb{P}^3, \mathcal{I}(d_2)) \neq H^0(\mathbb{P}^3, \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)),$$ where $\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}$ is the ideal sheaf of $\mathcal{C}_0 \cup \mathcal{C}_i$. Since the intersection $\mathcal{C}_0 \cap \mathcal{C}_i$ has the right codimension (i.e. $3$), we have an exact sequence (see Sasha's answer to Products of Ideal Sheaves and Union of irreducible Subvarieties):

$$0 \longrightarrow \mathcal{I}_{\mathcal{C}_0 \cup \mathcal{C}_i} \longrightarrow \mathcal{O}_{\mathbb{P}^3} \longrightarrow \mathcal{O}_{\mathcal{C}_0} \oplus \mathcal{O}_{\mathcal{C}_i} \longrightarrow \mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i} \longrightarrow 0.$$

Tensoring by $\mathcal{O}_{\mathbb{P}^3}(d_2)$ and taking Euler carateristics, we get: $$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{O}_{\mathbb{P}^3}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_i}(d_2)) + \chi(\mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ that is:

$$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ where $\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}$ is the ideal sheaf of $\mathcal{C}_0 \cap \mathcal{C}_i$ in $\mathcal{O}_{\mathcal{C}_i}$.

Applying Serre's vanishing theorem one more time, we find that for $d_2\gg0$:

  • $\chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = H^0(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2))$, $\chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) = H^0(\mathcal{J}_{\mathcal{C}_0}(d_2))$ and $\chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) = H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2))$,

  • $H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) \neq 0$.

As a consequence, for $d_2(i)\gg0$, the generic hypersurface of degree $d_2(i)$ containing $\mathcal{C}_0$ does not contain $\mathcal{C}_i$. This is true for all $i$ and since $k$ is infinite, by taking $d_3$ the max of the $d_2(i)$, we find that the generic hypersurface of degree $d_3$ does not contain any of the $\mathcal{C}_i$ for $i \neq 0$. Let $S_3$ be such a generic hypersurface, then we have, set-theoretically:

$$S_1 \cap S_2 \cap S_3 = \mathcal{C}_0.$$

Edit added following comments below tbe answer.
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Libli
Libli
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This is not precisely an answer to your question (since you would like yo understand Kneser's paper), but at the expense of using some modern technology (mainly sheaf cohomology) and being a little less constructive, I think one can get a very short proof of the result you mention.

Let $\mathcal{C} \subset \mathbb{P}^{3}$ be an integral projective curve (over an algebraically closed field) and denote by $\mathcal{I}$ its sheaf of ideal in $\mathbb{P}^3$. For any $d \geq 0$, the vector space:

$$H^0(\mathbb{P}^3, \mathcal{I}(d))$$

is the vector space of hypersurfaces of degree $d$ which contain $\mathcal{C}$. By Serre's vanishing Theorem, we know that the sheaf $\mathcal{I}(d_1)$ is globally generated for $d_1>>0$, in particular $H^0(\mathbb{P}^3, \mathcal{I}(d_1)) \neq 0$ for $d_1>>0$. An explicit bound for $d_1$ should be equivalent to the computation of the degree of your $f$ generating $I \cap k[x,y,z]$.

Let $f_1, f_2$ be two generic elements of $H^0(\mathbb{P}^3, \mathcal{I}(d_1))$. Up to increasing $d_1$, we can assume that the hypersurfaces $S_1$ and $S_2$ given by $f_1 = 0$ and $f_2 =0$ are integral. So in particular, we have, set-theoretically:

$$ S_1 \cap S_2 = \bigcup_{i=0}^{r} \mathcal{C}_i,$$ with $\mathcal{C}_0 = \mathcal{C}$ and the $\mathcal{C}_i$ are integral projective curves that meet properly.

Now, it is sufficient to find $d_2>0$ such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2))$ does not contain any of the $\mathcal{C}_i$ for $i>0$. Since $k$ is infinite and any vector space over $k$ is thus infinte, it is sufficient to prove that given a fixed $i \neq 0$, there exists $d_2(i) >0$, such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2(i)))$ does not contain $\mathcal{C}_i$ (and then apply the result for all $i$ by taking the complement in $H^0(\mathbb{P}^3, \mathcal{I}(d_3))$ of the union of the subspaces to be avoided, where $d_3$ is the max of the $d_2(i)$).

Thus, we are left to prove that for $d_2>>0$, we have: $$ H^0(\mathbb{P}^3, \mathcal{I}(d_2)) \neq H^0(\mathbb{P}^3, \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)),$$ where $\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}$ is the ideal sheaf of $\mathcal{C}_0 \cup \mathcal{C}_i$. Since the intersection $\mathcal{C}_0 \cap \mathcal{C}_i$ has the right codimension (i.e. $3$), we have an exact sequence (see the first answer to this question):

$$0 \longrightarrow \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}} \longrightarrow \mathcal{O}_{\mathbb{P}^3} \longrightarrow \mathcal{O}_{\mathcal{C}_0} \oplus \mathcal{O}_{\mathcal{C}_i} \longrightarrow \mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i} \longrightarrow 0.$$

Tensoring by $\mathcal{O}_{\mathbb{P}^3}(d_2)$ and taking Euler carateristics, we get: $$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{O}_{\mathbb{P}^3}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_i}(d_2)) + \chi(\mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ that is:

$$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ where $\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}$ is the ideal sheaf of $\mathcal{C}_0 \cap \mathcal{C}_i$ in $\mathcal{O}_{\mathcal{C}_i}$.

Applying Serre's vanishing theorem one more time, we find that for $d_2>>0$:

$\bullet$ $\chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = H^0(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2))$, $\chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) = H^0(\mathcal{J}_{\mathcal{C}_0}(d_2))$ and $\chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) = H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2))$,

$\bullet$ $H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) \neq 0$.

As a consequence, for $d_2(i)>>0$, the generic hypersurface of degree $d_2(i)$ containing $\mathcal{C}_0$ does not contain $\mathcal{C}_i$. This is true for all $i$ and since $k$ is infinite, by taking $d_3$ the max of the $d_2(i)$, we find that the generic hypersurface of degree $d_3$ does not contain any of the $\mathcal{C}_i$ for $i \neq 0$. Let $S_3$ be such a generic hypersurface, then we have, set-theoretically:

$$S_1 \cap S_2 \cap S_3 = \mathcal{C}_0.$$

EDIT : as abx points out in the comment below, my proof does not quite work. A fourth hypersurface containing $C_0$ of sufficiently high degree is needed to get rid off the potentiel points of intersection of $S_3$ with the $C_i$ ($i>0$) which do not lie on $C_0$.

This is not precisely an answer to your question (since you would like yo understand Kneser's paper), but at the expense of using some modern technology (mainly sheaf cohomology) and being a little less constructive, I think one can get a very short proof of the result you mention.

Let $\mathcal{C} \subset \mathbb{P}^{3}$ be an integral projective curve (over an algebraically closed field) and denote by $\mathcal{I}$ its sheaf of ideal in $\mathbb{P}^3$. For any $d \geq 0$, the vector space:

$$H^0(\mathbb{P}^3, \mathcal{I}(d))$$

is the vector space of hypersurfaces of degree $d$ which contain $\mathcal{C}$. By Serre's vanishing Theorem, we know that the sheaf $\mathcal{I}(d_1)$ is globally generated for $d_1>>0$, in particular $H^0(\mathbb{P}^3, \mathcal{I}(d_1)) \neq 0$ for $d_1>>0$. An explicit bound for $d_1$ should be equivalent to the computation of the degree of your $f$ generating $I \cap k[x,y,z]$.

Let $f_1, f_2$ be two generic elements of $H^0(\mathbb{P}^3, \mathcal{I}(d_1))$. Up to increasing $d_1$, we can assume that the hypersurfaces $S_1$ and $S_2$ given by $f_1 = 0$ and $f_2 =0$ are integral. So in particular, we have, set-theoretically:

$$ S_1 \cap S_2 = \bigcup_{i=0}^{r} \mathcal{C}_i,$$ with $\mathcal{C}_0 = \mathcal{C}$ and the $\mathcal{C}_i$ are integral projective curves that meet properly.

Now, it is sufficient to find $d_2>0$ such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2))$ does not contain any of the $\mathcal{C}_i$ for $i>0$. Since $k$ is infinite and any vector space over $k$ is thus infinte, it is sufficient to prove that given a fixed $i \neq 0$, there exists $d_2(i) >0$, such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2(i)))$ does not contain $\mathcal{C}_i$ (and then apply the result for all $i$ by taking the complement in $H^0(\mathbb{P}^3, \mathcal{I}(d_3))$ of the union of the subspaces to be avoided, where $d_3$ is the max of the $d_2(i)$).

Thus, we are left to prove that for $d_2>>0$, we have: $$ H^0(\mathbb{P}^3, \mathcal{I}(d_2)) \neq H^0(\mathbb{P}^3, \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)),$$ where $\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}$ is the ideal sheaf of $\mathcal{C}_0 \cup \mathcal{C}_i$. Since the intersection $\mathcal{C}_0 \cap \mathcal{C}_i$ has the right codimension (i.e. $3$), we have an exact sequence (see the first answer to this question):

$$0 \longrightarrow \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}} \longrightarrow \mathcal{O}_{\mathbb{P}^3} \longrightarrow \mathcal{O}_{\mathcal{C}_0} \oplus \mathcal{O}_{\mathcal{C}_i} \longrightarrow \mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i} \longrightarrow 0.$$

Tensoring by $\mathcal{O}_{\mathbb{P}^3}(d_2)$ and taking Euler carateristics, we get: $$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{O}_{\mathbb{P}^3}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_i}(d_2)) + \chi(\mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ that is:

$$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ where $\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}$ is the ideal sheaf of $\mathcal{C}_0 \cap \mathcal{C}_i$ in $\mathcal{O}_{\mathcal{C}_i}$.

Applying Serre's vanishing theorem one more time, we find that for $d_2>>0$:

$\bullet$ $\chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = H^0(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2))$, $\chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) = H^0(\mathcal{J}_{\mathcal{C}_0}(d_2))$ and $\chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) = H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2))$,

$\bullet$ $H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) \neq 0$.

As a consequence, for $d_2(i)>>0$, the generic hypersurface of degree $d_2(i)$ containing $\mathcal{C}_0$ does not contain $\mathcal{C}_i$. This is true for all $i$ and since $k$ is infinite, by taking $d_3$ the max of the $d_2(i)$, we find that the generic hypersurface of degree $d_3$ does not contain any of the $\mathcal{C}_i$ for $i \neq 0$. Let $S_3$ be such a generic hypersurface, then we have, set-theoretically:

$$S_1 \cap S_2 \cap S_3 = \mathcal{C}_0.$$

This is not precisely an answer to your question (since you would like yo understand Kneser's paper), but at the expense of using some modern technology (mainly sheaf cohomology) and being a little less constructive, I think one can get a very short proof of the result you mention.

Let $\mathcal{C} \subset \mathbb{P}^{3}$ be an integral projective curve (over an algebraically closed field) and denote by $\mathcal{I}$ its sheaf of ideal in $\mathbb{P}^3$. For any $d \geq 0$, the vector space:

$$H^0(\mathbb{P}^3, \mathcal{I}(d))$$

is the vector space of hypersurfaces of degree $d$ which contain $\mathcal{C}$. By Serre's vanishing Theorem, we know that the sheaf $\mathcal{I}(d_1)$ is globally generated for $d_1>>0$, in particular $H^0(\mathbb{P}^3, \mathcal{I}(d_1)) \neq 0$ for $d_1>>0$. An explicit bound for $d_1$ should be equivalent to the computation of the degree of your $f$ generating $I \cap k[x,y,z]$.

Let $f_1, f_2$ be two generic elements of $H^0(\mathbb{P}^3, \mathcal{I}(d_1))$. Up to increasing $d_1$, we can assume that the hypersurfaces $S_1$ and $S_2$ given by $f_1 = 0$ and $f_2 =0$ are integral. So in particular, we have, set-theoretically:

$$ S_1 \cap S_2 = \bigcup_{i=0}^{r} \mathcal{C}_i,$$ with $\mathcal{C}_0 = \mathcal{C}$ and the $\mathcal{C}_i$ are integral projective curves that meet properly.

Now, it is sufficient to find $d_2>0$ such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2))$ does not contain any of the $\mathcal{C}_i$ for $i>0$. Since $k$ is infinite and any vector space over $k$ is thus infinte, it is sufficient to prove that given a fixed $i \neq 0$, there exists $d_2(i) >0$, such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2(i)))$ does not contain $\mathcal{C}_i$ (and then apply the result for all $i$ by taking the complement in $H^0(\mathbb{P}^3, \mathcal{I}(d_3))$ of the union of the subspaces to be avoided, where $d_3$ is the max of the $d_2(i)$).

Thus, we are left to prove that for $d_2>>0$, we have: $$ H^0(\mathbb{P}^3, \mathcal{I}(d_2)) \neq H^0(\mathbb{P}^3, \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)),$$ where $\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}$ is the ideal sheaf of $\mathcal{C}_0 \cup \mathcal{C}_i$. Since the intersection $\mathcal{C}_0 \cap \mathcal{C}_i$ has the right codimension (i.e. $3$), we have an exact sequence (see the first answer to this question):

$$0 \longrightarrow \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}} \longrightarrow \mathcal{O}_{\mathbb{P}^3} \longrightarrow \mathcal{O}_{\mathcal{C}_0} \oplus \mathcal{O}_{\mathcal{C}_i} \longrightarrow \mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i} \longrightarrow 0.$$

Tensoring by $\mathcal{O}_{\mathbb{P}^3}(d_2)$ and taking Euler carateristics, we get: $$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{O}_{\mathbb{P}^3}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_i}(d_2)) + \chi(\mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ that is:

$$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ where $\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}$ is the ideal sheaf of $\mathcal{C}_0 \cap \mathcal{C}_i$ in $\mathcal{O}_{\mathcal{C}_i}$.

Applying Serre's vanishing theorem one more time, we find that for $d_2>>0$:

$\bullet$ $\chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = H^0(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2))$, $\chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) = H^0(\mathcal{J}_{\mathcal{C}_0}(d_2))$ and $\chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) = H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2))$,

$\bullet$ $H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) \neq 0$.

As a consequence, for $d_2(i)>>0$, the generic hypersurface of degree $d_2(i)$ containing $\mathcal{C}_0$ does not contain $\mathcal{C}_i$. This is true for all $i$ and since $k$ is infinite, by taking $d_3$ the max of the $d_2(i)$, we find that the generic hypersurface of degree $d_3$ does not contain any of the $\mathcal{C}_i$ for $i \neq 0$. Let $S_3$ be such a generic hypersurface, then we have, set-theoretically:

$$S_1 \cap S_2 \cap S_3 = \mathcal{C}_0.$$

EDIT : as abx points out in the comment below, my proof does not quite work. A fourth hypersurface containing $C_0$ of sufficiently high degree is needed to get rid off the potentiel points of intersection of $S_3$ with the $C_i$ ($i>0$) which do not lie on $C_0$.

Fix some indexes typos and spelling mistakes
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This is not precisely an answer to your question (since you would like yo understand Kneser's paper), but at the expense of using some modern technologiestechnology (mainly sheaf cohomology) and being a little less constructive, I think one can get a very short proof of the result you mention. I will also mentio explicit bounds and constructive methods when possible.

Let $\mathcal{C} \subset \mathbb{P}^{3}$ be an integral projective curve (over an algebraically closed field) and denote by $\mathcal{I}$ its sheaf of ideal in $\mathbb{P}^3$. For any $d \geq 0$, the vector space:

$$H^0(\mathbb{P}^3, \mathcal{I}(d))$$

is the vector space of hyeprsurfacehypersurfaces of degree $d$ which containscontain $\mathcal{C}$. By Serre's vanishing Theorem, we know that the sheaf $\mathcal{I}(d)$$\mathcal{I}(d_1)$ is globally generated for $d_1>>0$, in particular $H^0(\mathbb{P}^3, \mathcal{I}(d_1)) \neq 0$ for $d>>0$$d_1>>0$. An explicit bound for $d$$d_1$ should be equivalent to the computation of the degree of your $f$ generating $I \cap k[x,y,z]$.

Let $f_1, f_2$ be two generic elements of $H^0(\mathbb{P}^3, \mathcal{I}(d_1))$. Up to increasing $d_1$, wwewe can assume that the hypersurfaces $S_1$ and $S_2$ given by $f_1 = 0$ and $f_2 =0$ are integral. So in particular, we have, set-theoretically:

$$ S_1 \cap S_2 = \bigcup_{i=0}^{r} \mathcal{C}_i,$$ with $\mathcal{C}_0 = \mathcal{C}$ and the $\mathcal{C}_i$ are integral projective curves that meet properly.

Now, it is sufficient to find $d_2>0$ such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2))$ does not contain any of the $\mathcal{C}_i$ for $i>0$. Since $k$ is infinite and any vector space over $k$ is thus infinte, it is sufficient to prove that given a fixed $i \neq 0$, there exists $d_2(i) >0$, such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2(i)))$ does not contain $\mathcal{C}_i$ (and then apply the result for all $i$ by taking the complement in $H^0(\mathbb{P}^3, \mathcal{I}(d_3))$ of the union of the subspaces to be avoided, where $d_3$ is the max of the $d_2(i)$).

Thus, we are left to prove that for $d_2>>0$, we have: $$ H^0(\mathbb{P}^3, \mathcal{I}(d_2)) \neq H^0(\mathbb{P}^3, \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)),$$ where $\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}$ is the ideal sheaf of $\mathcal{C}_0 \cup \mathcal{C}_1$$\mathcal{C}_0 \cup \mathcal{C}_i$. Since the intersection $\mathcal{C}_0 \cap \mathcal{C}_1$$\mathcal{C}_0 \cap \mathcal{C}_i$ has the right codimension (i.e. $0$$3$), we have an exact sequence (see the first answer to this question):

$$0 \longrightarrow \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}} \longrightarrow \mathcal{O}_{\mathbb{P}^3} \longrightarrow \mathcal{O}_{\mathcal{C}_0} \oplus \mathcal{O}_{\mathcal{C}_1} \longrightarrow \mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_1} \longrightarrow 0.$$$$0 \longrightarrow \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}} \longrightarrow \mathcal{O}_{\mathbb{P}^3} \longrightarrow \mathcal{O}_{\mathcal{C}_0} \oplus \mathcal{O}_{\mathcal{C}_i} \longrightarrow \mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i} \longrightarrow 0.$$

Tensoring by $\mathcal{O}_{\mathbb{P}^3}(d_2)$ and taking Euler carateristiccarateristics, we get: $$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{O}_{\mathbb{P}^3}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_1}(d_2)) + \chi(\mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_1}(d_2)),$$$$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{O}_{\mathbb{P}^3}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_i}(d_2)) + \chi(\mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ that is:

$$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{J}^{\mathcal{C}_1}_{\mathcal{C}_0 \cap \mathcal{C}_1}(d_2)),$$$$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ where $\mathcal{J}^{\mathcal{C}_1}_{\mathcal{C}_0 \cap \mathcal{C}_1}$$\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}$ is the ideal sheaf of $\mathcal{C}_0 \cap \mathcal{C}_1$$\mathcal{C}_0 \cap \mathcal{C}_i$ in $\mathcal{O}_{\mathcal{C}_1}$$\mathcal{O}_{\mathcal{C}_i}$.

ApplyyingApplying Serre's vanishngvanishing theorem one more time, we find that for $d_2>>0$:

$\bullet$ $\chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = H^0(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2))$, $\chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) = H^0(\mathcal{J}_{\mathcal{C}_0}(d_2))$ and $\chi(\mathcal{J}^{\mathcal{C}_1}_{\mathcal{C}_0 \cap \mathcal{C}_1}(d_2)) = H^0(\mathcal{J}^{\mathcal{C}_1}_{\mathcal{C}_0 \cap \mathcal{C}_1}(d_2))$$\chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) = H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2))$,

$\bullet$ $H^0(\mathcal{J}^{\mathcal{C}_1}_{\mathcal{C}_0 \cap \mathcal{C}_1}(d_2)) \neq 0$$H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) \neq 0$.

As a consequence, for $d_2(i)>>0$, the generic hypersurface of degree $d_2(i)$ containing $\mathcal{C}_0$ does not contain $\mathcal{C}_i$. This is true for all $i$ and since $k$ is infinite, by taking $d_3$ the max of the $d_2(i)$, we find that the generic hypersurface of degree $d_3$ does not contain any of the $\mathcal{C}_i$ for $i \neq 0$. Let $S_3$ be such a generic hypersurface, then we have, set-theoretically:

$$S_1 \cap S_2 \cap S_3 = \mathcal{C}_0.$$

This is not precisely an answer to your question (since you would like yo understand Kneser's paper), but at the expense of using some modern technologies (mainly sheaf cohomology) and being a little less constructive, I think one can get a very short proof of the result you mention. I will also mentio explicit bounds and constructive methods when possible.

Let $\mathcal{C} \subset \mathbb{P}^{3}$ be an integral projective curve (over an algebraically closed field) and denote by $\mathcal{I}$ its sheaf of ideal in $\mathbb{P}^3$. For any $d \geq 0$, the vector space:

$$H^0(\mathbb{P}^3, \mathcal{I}(d))$$

is the vector space of hyeprsurface of degree $d$ which contains $\mathcal{C}$. By Serre's vanishing Theorem, we know that the sheaf $\mathcal{I}(d)$ is globally generated for $d_1>>0$, in particular $H^0(\mathbb{P}^3, \mathcal{I}(d_1)) \neq 0$ for $d>>0$. An explicit bound for $d$ should be equivalent to the computation of the degree of your $f$ generating $I \cap k[x,y,z]$.

Let $f_1, f_2$ be two generic elements of $H^0(\mathbb{P}^3, \mathcal{I}(d_1))$. Up to increasing $d_1$, wwe can assume that the hypersurfaces $S_1$ and $S_2$ given by $f_1 = 0$ and $f_2 =0$ are integral. So in particular, we have, set-theoretically:

$$ S_1 \cap S_2 = \bigcup_{i=0}^{r} \mathcal{C}_i,$$ with $\mathcal{C}_0 = \mathcal{C}$ and the $\mathcal{C}_i$ are integral projective curves that meet properly.

Now it is sufficient to find $d_2>0$ such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2))$ does not contain any of the $\mathcal{C}_i$ for $i>0$. Since $k$ is infinite and any vector space over $k$ is thus infinte, it is sufficient to prove that given a fixed $i \neq 0$, there exists $d_2(i) >0$, such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2(i)))$ does not contain $\mathcal{C}_i$ (and then apply the result for all $i$ by taking the complement in $H^0(\mathbb{P}^3, \mathcal{I}(d_3))$ of the union of the subspaces to be avoided, where $d_3$ is the max of the $d_2(i)$).

Thus we are left to prove that for $d_2>>0$, we have: $$ H^0(\mathbb{P}^3, \mathcal{I}(d_2)) \neq H^0(\mathbb{P}^3, \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)),$$ where $\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}$ is the ideal sheaf of $\mathcal{C}_0 \cup \mathcal{C}_1$. Since the intersection $\mathcal{C}_0 \cap \mathcal{C}_1$ has the right codimension (i.e. $0$), we have an exact sequence (see the first answer to this question):

$$0 \longrightarrow \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}} \longrightarrow \mathcal{O}_{\mathbb{P}^3} \longrightarrow \mathcal{O}_{\mathcal{C}_0} \oplus \mathcal{O}_{\mathcal{C}_1} \longrightarrow \mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_1} \longrightarrow 0.$$

Tensoring by $\mathcal{O}_{\mathbb{P}^3}(d_2)$ and taking Euler carateristic, we get: $$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{O}_{\mathbb{P}^3}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_1}(d_2)) + \chi(\mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_1}(d_2)),$$ that is:

$$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{J}^{\mathcal{C}_1}_{\mathcal{C}_0 \cap \mathcal{C}_1}(d_2)),$$ where $\mathcal{J}^{\mathcal{C}_1}_{\mathcal{C}_0 \cap \mathcal{C}_1}$ is the ideal sheaf of $\mathcal{C}_0 \cap \mathcal{C}_1$ in $\mathcal{O}_{\mathcal{C}_1}$.

Applyying Serre's vanishng theorem one more time, we find that for $d_2>>0$:

$\bullet$ $\chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = H^0(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2))$, $\chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) = H^0(\mathcal{J}_{\mathcal{C}_0}(d_2))$ and $\chi(\mathcal{J}^{\mathcal{C}_1}_{\mathcal{C}_0 \cap \mathcal{C}_1}(d_2)) = H^0(\mathcal{J}^{\mathcal{C}_1}_{\mathcal{C}_0 \cap \mathcal{C}_1}(d_2))$,

$\bullet$ $H^0(\mathcal{J}^{\mathcal{C}_1}_{\mathcal{C}_0 \cap \mathcal{C}_1}(d_2)) \neq 0$.

As a consequence, for $d_2(i)>>0$, the generic hypersurface of degree $d_2(i)$ containing $\mathcal{C}_0$ does not contain $\mathcal{C}_i$. This is true for all $i$ and since $k$ is infinite, by taking $d_3$ the max of the $d_2(i)$, we find that the generic hypersurface of degree $d_3$ does not contain any of the $\mathcal{C}_i$ for $i \neq 0$. Let $S_3$ be such a generic hypersurface, then we have, set-theoretically:

$$S_1 \cap S_2 \cap S_3 = \mathcal{C}_0.$$

This is not precisely an answer to your question (since you would like yo understand Kneser's paper), but at the expense of using some modern technology (mainly sheaf cohomology) and being a little less constructive, I think one can get a very short proof of the result you mention.

Let $\mathcal{C} \subset \mathbb{P}^{3}$ be an integral projective curve (over an algebraically closed field) and denote by $\mathcal{I}$ its sheaf of ideal in $\mathbb{P}^3$. For any $d \geq 0$, the vector space:

$$H^0(\mathbb{P}^3, \mathcal{I}(d))$$

is the vector space of hypersurfaces of degree $d$ which contain $\mathcal{C}$. By Serre's vanishing Theorem, we know that the sheaf $\mathcal{I}(d_1)$ is globally generated for $d_1>>0$, in particular $H^0(\mathbb{P}^3, \mathcal{I}(d_1)) \neq 0$ for $d_1>>0$. An explicit bound for $d_1$ should be equivalent to the computation of the degree of your $f$ generating $I \cap k[x,y,z]$.

Let $f_1, f_2$ be two generic elements of $H^0(\mathbb{P}^3, \mathcal{I}(d_1))$. Up to increasing $d_1$, we can assume that the hypersurfaces $S_1$ and $S_2$ given by $f_1 = 0$ and $f_2 =0$ are integral. So in particular, we have, set-theoretically:

$$ S_1 \cap S_2 = \bigcup_{i=0}^{r} \mathcal{C}_i,$$ with $\mathcal{C}_0 = \mathcal{C}$ and the $\mathcal{C}_i$ are integral projective curves that meet properly.

Now, it is sufficient to find $d_2>0$ such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2))$ does not contain any of the $\mathcal{C}_i$ for $i>0$. Since $k$ is infinite and any vector space over $k$ is thus infinte, it is sufficient to prove that given a fixed $i \neq 0$, there exists $d_2(i) >0$, such that the vanishing locus of a generic section in $H^0(\mathbb{P}^3, \mathcal{I}(d_2(i)))$ does not contain $\mathcal{C}_i$ (and then apply the result for all $i$ by taking the complement in $H^0(\mathbb{P}^3, \mathcal{I}(d_3))$ of the union of the subspaces to be avoided, where $d_3$ is the max of the $d_2(i)$).

Thus, we are left to prove that for $d_2>>0$, we have: $$ H^0(\mathbb{P}^3, \mathcal{I}(d_2)) \neq H^0(\mathbb{P}^3, \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)),$$ where $\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}$ is the ideal sheaf of $\mathcal{C}_0 \cup \mathcal{C}_i$. Since the intersection $\mathcal{C}_0 \cap \mathcal{C}_i$ has the right codimension (i.e. $3$), we have an exact sequence (see the first answer to this question):

$$0 \longrightarrow \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}} \longrightarrow \mathcal{O}_{\mathbb{P}^3} \longrightarrow \mathcal{O}_{\mathcal{C}_0} \oplus \mathcal{O}_{\mathcal{C}_i} \longrightarrow \mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i} \longrightarrow 0.$$

Tensoring by $\mathcal{O}_{\mathbb{P}^3}(d_2)$ and taking Euler carateristics, we get: $$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{O}_{\mathbb{P}^3}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_i}(d_2)) + \chi(\mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ that is:

$$ \chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ where $\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}$ is the ideal sheaf of $\mathcal{C}_0 \cap \mathcal{C}_i$ in $\mathcal{O}_{\mathcal{C}_i}$.

Applying Serre's vanishing theorem one more time, we find that for $d_2>>0$:

$\bullet$ $\chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = H^0(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2))$, $\chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) = H^0(\mathcal{J}_{\mathcal{C}_0}(d_2))$ and $\chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) = H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2))$,

$\bullet$ $H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) \neq 0$.

As a consequence, for $d_2(i)>>0$, the generic hypersurface of degree $d_2(i)$ containing $\mathcal{C}_0$ does not contain $\mathcal{C}_i$. This is true for all $i$ and since $k$ is infinite, by taking $d_3$ the max of the $d_2(i)$, we find that the generic hypersurface of degree $d_3$ does not contain any of the $\mathcal{C}_i$ for $i \neq 0$. Let $S_3$ be such a generic hypersurface, then we have, set-theoretically:

$$S_1 \cap S_2 \cap S_3 = \mathcal{C}_0.$$

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