First of all, you definitely need to assume that your curve is proper to make sense of $\chi_k(\mathcal O_X)$. If it is not, then $H^i(X,\mathcal O_X)$ is not finite dimensional over $k$, so $\chi_X(\mathcal O_X)$ is not well-defined. Actually, it is true that any proper curve is projective (Hartshorne, exercise III.5.8). Thus it is harmless to assume that $X$ is projective.

Let me now deal with your first question:

Proposition 1: Let X be a proper integral scheme with a $k$-rational point $a\in X(k)$, then $H^0(X,\mathcal O_X)=k$.

Proof: Properness of $X$ over $k$ guarantees that $A:=H^0(X,\mathcal O_X)$ is a finite dimensional algebra over $k$. Any finite dimensional commutative algebra over $k$ is equal to a product of local finite dimensional $k$-algebras. In other words, $A=\prod_i A_i$ where each $A_i$ is a local finite dimensional $k$-algebra. Since $X$ is connected we conclude that $i=1$, otherwise we will have idempotents inside $H^0(X,\mathcal O_{X})=A$. Therefore, $A$ is actually a local finite dimensional $k$-algebra. Reducedness of $X$ implies that $A$ should also be reduced. Any finite-dimensional reduced algebra over a field is a field $k'$ and an extension $k \subset k'$ should be finite. Observe that there is a natural morphism $f:X \to \operatorname{Spec} H^0(X,\mathcal O_X)=\operatorname{Spec} k'.$ And a composition of $f$ with a section $a:\operatorname{Spec} k \to X$ defines a section $a':\operatorname{Spec} k \to \operatorname{Spec} k'$. Now it is easy to conclude that $k=k'$.

Finally, it is not true without any regularity hypothesis that $\mathcal O_X(a)$ is invertible for an arbitrary $a\in X(k)$. In order to have this property you need to assume that $a$ is a regular point of $X$. However, I will not assume this and will prove a little bit more general result. Namely, I am going to prove the following statement:

Propostion 2: Let $X$ be a integral curve of arithmetic genus $0$ over a field $k$. Assume that $H^0(X,\mathcal O_X)=k$ and let $\mathcal L$ be an invertible sheaf with a non-zero section $s\in H^0(X,\mathcal L)$, then $\mathcal L$ is globally generated.

Proof: Choose any section $s\in H^0(X,\mathcal L)$ and let $D:=V(s)$ be a corresponding Cartier divisor. Basically by definition we have $\mathcal L=\mathcal O_X(D)$. Note that for any $x\notin D$ we know that $s|_{x}\neq 0$. Thus $s$ generates $\mathcal L$ outside of $D$.

Now let us show that for any $x\in D$ there is a section $t\in H^0(X, \mathcal L)$ such that $t|_{x}\neq 0$. Recall that we have a short exact sequence

$$ 0 \to \mathcal O_X \to \mathcal O_X(D) \to \mathcal O_D \to 0 (*). $$

Since $D$ is a zero-dimensional scheme over a field, we see that $\mathcal O_D$ is globally generated. In particular, there is a section $t'\in H^0(X, \mathcal O_D)$ such that $t'|_{x}\neq 0$. Hence, the only thing we are left to show is that the natural morphism $$ \phi:H^0 (X, \mathcal O_X(D)) \to H^0(X, \mathcal O_D) $$ is surjective. It suffices to show that $H^1(X, \mathcal O_X)=0$ (because of the long exact sequence associated to $(*)$). However, by the very definition of arithmetic genus we know that $$ 0=p_a(X)=1-\chi_X(\mathcal O_X)=1-H^0(X,\mathcal O_X)+H^1(X,\mathcal O_X)=H^1(X, \mathcal O_X). $$

Therefore, the map $\phi$ is indeed surjective. Thus there is $t\in H^0(X,\mathcal O_X(D))$ such that $\phi(t)=t'$. Hence $t|_{x}=t'|_{x} \neq 0$. So, $\mathcal L = \mathcal O_X(D)$ is generated by global sections at $x$. Since $x$ was an arbitrary point of $D$, we conclude that $\mathcal L$ is generated by global sections at any point of $D$. As a result, it is globally generated.

First of all, you definitely need to assume that your curve is proper to make sense of $\chi_k(\mathcal O_X)$. If it is not, then $H^i(X,\mathcal O_X)$ is not finite dimensional over $k$, so $\chi_X(\mathcal O_X)$ is not well-defined. Actually, it is true that any proper curve is projective (Hartshorne, exercise III.5.8). Thus it is harmless to assume that $X$ is projective.

Let me now deal with your first question:

Proposition 1: Let X be a proper integral scheme with a $k$-rational point $a\in X(k)$, then $H^0(X,\mathcal O_X)=k$.

Proof: Properness of $X$ over $k$ guarantees that $A:=H^0(X,\mathcal O_X)$ is a finite dimensional algebra over $k$. Any finite dimensional commutative algebra over $k$ is equal to a product of local finite dimensional $k$-algebras. In other words, $A=\prod_i A_i$ where each $A_i$ is a local finite dimensional $k$-algebra. Since $X$ is connected we conclude that $i=1$, otherwise we will have idempotents inside $H^0(X,\mathcal O_{X})=A$. Therefore, $A$ is actually a local finite dimensional $k$-algebra. Reducedness of $X$ implies that $A$ should also be reduced. Any finite-dimensional reduced algebra over a field is a field $k'$ and an extension $k \subset k'$ should be finite. Observe that there is a natural morphism $f:X \to \operatorname{Spec} H^0(X,\mathcal O_X)=\operatorname{Spec} k'.$ And a composition of $f$ with a section $a:\operatorname{Spec} k \to X$ defines a section $a':\operatorname{Spec} k \to \operatorname{Spec} k'$. Now it is easy to conclude that $k=k'$.

Finally, it is not true without any regularity hypothesis that $\mathcal O_X(a)$ is invertible for an arbitrary $a\in X(k)$. In order to have this property you need to assume that $a$ is a regular point of $X$. However, I will not assume this and will prove a little bit more general result. Namely, I am going to prove the following statement:

Propostion 2: Let $X$ be an integral curve of arithmetic genus $0$ over a field $k$. Assume that $H^0(X,\mathcal O_X)=k$ and let $\mathcal L$ be an invertible sheaf with a non-zero section $s\in H^0(X,\mathcal L)$, then $\mathcal L$ is globally generated.

Proof: Choose any section $s\in H^0(X,\mathcal L)$ and let $D:=V(s)$ be a corresponding Cartier divisor. Basically by definition we have $\mathcal L=\mathcal O_X(D)$. Note that for any $x\notin D$ we know that $s|_{x}\neq 0$. Thus $s$ generates $\mathcal L$ outside of $D$.

Now let us show that for any $x\in D$ there is a section $t\in H^0(X, \mathcal L)$ such that $t|_{x}\neq 0$. Recall that we have a short exact sequence

$$ 0 \to \mathcal O_X \to \mathcal O_X(D) \to \mathcal O_D \to 0 (*). $$

Since $D$ is a zero-dimensional scheme over a field, we see that $\mathcal O_D$ is globally generated. In particular, there is a section $t'\in H^0(X, \mathcal O_D)$ such that $t'|_{x}\neq 0$. Hence, the only thing we are left to show is that the natural morphism $$ \phi:H^0 (X, \mathcal O_X(D)) \to H^0(X, \mathcal O_D) $$ is surjective. It suffices to show that $H^1(X, \mathcal O_X)=0$ (because of the long exact sequence associated to $(*)$). However, by the very definition of arithmetic genus we know that $$ 0=p_a(X)=1-\chi_X(\mathcal O_X)=1-H^0(X,\mathcal O_X)+H^1(X,\mathcal O_X)=H^1(X, \mathcal O_X). $$

Therefore, the map $\phi$ is indeed surjective. Thus there is $t\in H^0(X,\mathcal O_X(D))$ such that $\phi(t)=t'$. Hence $t|_{x}=t'|_{x} \neq 0$. So, $\mathcal L = \mathcal O_X(D)$ is generated by global sections at $x$. Since $x$ was an arbitrary point of $D$, we conclude that $\mathcal L$ is generated by global sections at any point of $D$. As a result, it is globally generated.

Secondly, you need to assume that your curve is geometrically integral to prove that $H^0(X,\mathcal O_X)=k$. Since you had problems to show it by yourself, let me explain it here. Note that properness of $X$ over $k$ guarantees that $A:=H^0(X,\mathcal O_X)$ is a finite dimensional algebra over $k$. By flat base change (apply it to a flat morphism Spec $\bar k \to $ Spec $k$) we obtain $H^0(X\otimes_k \bar k,\mathcal O_{X\otimes_k \bar k})=A\otimes_k \bar k$. Any finite dimensional commutative algebra over $\bar k$ is equal to a product of local finite dimensional $\bar k$-algebras. In other words, $A\otimes_k \bar k=\prod_i A_i$ where each $A_i$ is a local finite dimensional $\bar k$-algebra. Since $X\otimes_k \bar k$ is connected we conclude that $i=1$, otherwise we will have idempotents inside $H^0(X\otimes_k \bar k,\mathcal O_{X\otimes_k \bar k})=A\otimes_k \bar k$. Therefore, $A\otimes_k \bar k$ is actually a local finite dimensional $\bar k$-algebra. Note that also we require $X\otimes_k \bar k$ to be reduced, thus we know that $A\otimes_k \bar k$ should also be reduced. But the only one local finite dimensional $\bar k$-algebra is $\bar k$ itself (exercise). So, we conclude that $A\otimes_k \bar k=\bar k$. As a consequence, $\dim_k A=\dim_{\bar k}A\otimes \bar k=1 =>A=k$.

Thirdly, it is not true without any regularity hypothesis that $\mathcal O_X(a)$ is invertible for any $a\in X(k)$. In order to have this property you need to assume that $a$ is a regular point of $X$. However, I will not assume this and will prove a little bit more general result. Namely, I am going to prove the following statement:

Let $X$ be a geometrically integral curve of arithmetic genus $0$ over a field $k$. And let $\mathcal L$ be an invertible sheaf with a non-zero section $s\in H^0(X,\mathcal L)$, then $\mathcal L$ is globally generated.

Let me now deal with your first question:

Proposition 1: Let X be a proper integral scheme with a $k$-rational point $a\in X(k)$, then $H^0(X,\mathcal O_X)=k$.

Proof: Properness of $X$ over $k$ guarantees that $A:=H^0(X,\mathcal O_X)$ is a finite dimensional algebra over $k$. Any finite dimensional commutative algebra over $k$ is equal to a product of local finite dimensional $k$-algebras. In other words, $A=\prod_i A_i$ where each $A_i$ is a local finite dimensional $k$-algebra. Since $X$ is connected we conclude that $i=1$, otherwise we will have idempotents inside $H^0(X,\mathcal O_{X})=A$. Therefore, $A$ is actually a local finite dimensional $k$-algebra. Reducedness of $X$ implies that $A$ should also be reduced. Any finite-dimensional reduced algebra over a field is a field $k'$ and an extension $k \subset k'$ should be finite. Observe that there is a natural morphism $f:X \to \operatorname{Spec} H^0(X,\mathcal O_X)=\operatorname{Spec} k'.$ And a composition of $f$ with a section $a:\operatorname{Spec} k \to X$ defines a section $a':\operatorname{Spec} k \to \operatorname{Spec} k'$. Now it is easy to conclude that $k=k'$.

Finally, it is not true without any regularity hypothesis that $\mathcal O_X(a)$ is invertible for an arbitrary $a\in X(k)$. In order to have this property you need to assume that $a$ is a regular point of $X$. However, I will not assume this and will prove a little bit more general result. Namely, I am going to prove the following statement:

Propostion 2: Let $X$ be a integral curve of arithmetic genus $0$ over a field $k$. Assume that $H^0(X,\mathcal O_X)=k$ and let $\mathcal L$ be an invertible sheaf with a non-zero section $s\in H^0(X,\mathcal L)$, then $\mathcal L$ is globally generated.