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Proof. Let $a$ be a point of $\ell \setminus A$: such a point exists, since the points of $A$ are not collinear and $\ell$ contains $q+1$ points. There are $q$ lines different from $\ell$ through $a$ and at most $q-2$ points in $A \setminus \ell$; thus there are two distinct lines $\ell_1 , \ell_2$ through $a$ disjoint from $A$. Let $b$ be any point of $A$ not contained in $\ell$; there are $q+1$ lines through $b$ and only $q$ points in $A$ different from $b$; we deduce that there is a line $\ell_3$ through $b$ such that $A \cap \ell_3 = \{b\}$. Thus the triangle formed by $\ell_1,\ell_2,\ell_3$ is a blocking set disjoint from $A$. $\square$

Proof. Let $a$ be a point of $\ell \setminus A$: such a point exists, since the points of $A$ are not collinear and $\ell$ contains $q+1$ points. There are $q$ lines different from $\ell$ through $a$ and at most $q-2$ points in $A \setminus \ell$; thus there are two distinct lines $\ell_1 , \ell_2$ through $a$ disjoint from $A$. Let $b'$ be any point of $A$ not contained in $\ell$ and let $b$ be any point not contained in $A$ on a line $\ell'$ joining $b'$ to a point of $A$ contained in $\ell$; there are $q$ lines through $b$ different from $\ell'$ and only $q-1$ points in $A \setminus \ell'$; we deduce that there is a line $\ell_3$ through $b$ disjoint from $A$. Thus the triangle formed by $\ell_1,\ell_2,\ell_3$ is a blocking set disjoint from $A$. $\square$

Let $\ell_1,\ell_2,\ell_3$ be three lines in the plane that do not all contain the same point. The triangle formed by $\ell_1,\ell_2,\ell_3$ is the set obtained from $\ell_1 \cup \ell_2 \cup \ell_3$ by removing the three pairwise intersections of the lines. Clearly, any triangle is a blocking set: every line in the plane meets each line $\ell_1,\ell_2,\ell_3$, and cannot contain all the three points that we are removing from the triangle. Triangles are the only blocking sets we consider.

We say that $A$ has property (*) if the intersection of $A$ with every triangle is not empty. Note that if a subset of the plane intersects non-trivially every blocking set, then it has property (*). We prove the (a priori) stronger statement that a set with property (*) and at most $q+1$ elements consists of all the points contained in a line.

Lemma 1. Let $A$ be a set of at least three points in the plane such that any line intersects $A$ in at most two points; then $A$ cannot have property (*).

Proof. Let $A' \subset A$ be a subset with three elements, and note that the points in $A'$ are not collinear. Let $\ell_1,\ell_2,\ell_3$ be the three lines each containing two of the points of $A'$. By assumption $\ell_1 \cap A , \ell_2 \cap A , \ell_3 \cap A \subset A'$; thus, the triangle formed by $\ell_1 , \ell_2 , \ell_3$ is a blocking set disjoint from $A$. $\square$

Lemma 2. Let $A$ be a set of at most $q+1$ points in the plane that are not collinear and let $\ell$ be a line in the plane such that $|A \cap \ell| \geq 3$; then $A$ cannot have property (*).

Proof. Let $a$ be a point of $\ell \setminus A$: such a point exists, since the points of $A$ are not collinear and $\ell$ contains $q+1$ points. There are $q$ lines different from $\ell$ through $a$ and at most $q-2$ points in $A \setminus \ell$; thus there are two distinct lines $\ell_1 , \ell_2$ through $a$ disjoint from $A$. Let $b$ be any point of $A$ not contained in $\ell$; there are $q+1$ lines through $b$ and only $q$ points in $A$ different from $b$; we deduce that there is a line $\ell_3$ through $b$ and disjoint from $A$. Thus the triangle formed by $\ell_1,\ell_2,\ell_3$ is a blocking set disjoint from $A$. $\square$

Corollary. Let $A$ be a set of at most $q+1$ points in the plane having property (*); then $A$ consists of all the points contained in a line.

Proof. By Lemma 1 we know that there must be a line containing three points of $A$. If the points of $A$ were not collinear, then Lemma 2 would imply that $A$ does not have property (*). Thus the points of $A$ are collinear. To conclude it suffices to show that if $A$ misses a point of the line containing $A$, then $A$ does not have property (*). This is easy: let $a$ be a point of $A$ and let $\ell_1,\ell_2$ be distinct lines through $a$; let also $b$ be a point on the line containing $A$ not in $A$, and let $\ell_3$ be any line through $b$ not containing $A$. The triangle formed by $\ell_1,\ell_2,\ell_3$ is a blocking set disjoint from $A$. $\square$

Let $\ell_1,\ell_2,\ell_3$ be three lines in the plane that do not all contain the same point. The triangle formed by $\ell_1,\ell_2,\ell_3$ is the set obtained from $\ell_1 \cup \ell_2 \cup \ell_3$ by removing the three pairwise intersections of the lines. Clearly, any triangle is a blocking set: every line in the plane meets each line $\ell_1,\ell_2,\ell_3$, and cannot contain all the three points that we are removing from the triangle. Triangles are the only blocking sets we consider.

We say that $A$ has property (*) if the intersection of $A$ with every triangle is not empty. Note that if a subset of the plane intersects non-trivially every blocking set, then it has property (*). We prove the (a priori) stronger statement that a set with property (*) and at most $q+1$ elements consists of all the points contained in a line.

Lemma 1. Let $A$ be a set of at least three points in the plane such that any line intersects $A$ in at most two points; then $A$ cannot have property (*).

Proof. Let $A' \subset A$ be a subset with three elements, and note that the points in $A'$ are not collinear. Let $\ell_1,\ell_2,\ell_3$ be the three lines each containing two of the points of $A'$. By assumption $\ell_1 \cap A , \ell_2 \cap A , \ell_3 \cap A \subset A'$; thus, the triangle formed by $\ell_1 , \ell_2 , \ell_3$ is a blocking set disjoint from $A$. $\square$

Lemma 2. Let $A$ be a set of at most $q+1$ points in the plane that are not collinear and let $\ell$ be a line in the plane such that $|A \cap \ell| \geq 3$; then $A$ cannot have property (*).

Proof. Let $a$ be a point of $\ell \setminus A$: such a point exists, since the points of $A$ are not collinear and $\ell$ contains $q+1$ points. There are $q$ lines different from $\ell$ through $a$ and at most $q-2$ points in $A \setminus \ell$; thus there are two distinct lines $\ell_1 , \ell_2$ through $a$ disjoint from $A$. Let $b$ be any point of $A$ not contained in $\ell$; there are $q+1$ lines through $b$ and only $q$ points in $A$ different from $b$; we deduce that there is a line $\ell_3$ through $b$ such that $A \cap \ell_3 = \{b\}$. Thus the triangle formed by $\ell_1,\ell_2,\ell_3$ is a blocking set disjoint from $A$. $\square$

Corollary. Let $A$ be a set of at most $q+1$ points in the plane having property (*); then $A$ consists of all the points contained in a line.

Proof. By Lemma 1 we know that there must be a line containing three points of $A$. If the points of $A$ were not collinear, then Lemma 2 would imply that $A$ does not have property (*). Thus the points of $A$ are collinear. To conclude it suffices to show that if $A$ misses a point of the line containing $A$, then $A$ does not have property (*). This is easy: let $a$ be a point of $A$ and let $\ell_1,\ell_2$ be distinct lines through $a$; let also $b$ be a point on the line containing $A$ not in $A$, and let $\ell_3$ be any line through $b$ not containing $A$. The triangle formed by $\ell_1,\ell_2,\ell_3$ is a blocking set disjoint from $A$. $\square$

I think that I can show the inequality, provided I understood the definitions and I did not make mistakes. If this strategy works, it can probably be argued in a more streamlined manner, but here it goes.

From what I understand of the definition of a blocking set, the following should be one. Fix a line $\ell$ in the plane, a point $p$ on $\ell$, and a conic $C$ containing $p$ and tangent to $\ell$ at $p$. The set of points of $\ell$ and $C$, without the point $p$ is a blocking set: any line not containing $p$, meets $\ell$ away from $p$; any line through $p$ either meets $C$ at a single point different from $p$ or coincides with $\ell$. Observe that for a given pair $(p,C)$ of a point $p$ and a conic $C$ containing $p$, there are exactly $q$ smooth conics, including $C$, having contact order four with $C$ at $p$. Any two distinct of these $q$ conics only have $p$ as a common point. In homogeneous coordinates $x,y,z$ on the plane, we can choose $C : \{x^2-yz=0\}$ and $p=[0,0,1]$ so that the mentioned conics are $C_\lambda : \{x^2+y(\lambda y - z) = 0 \}$, as $\lambda $ varies in the ground field; the common tangent line is the line $y=0$.

Case 1: there is a line $\ell$ in the plane such that $A \cap \ell$ consists of exactly one point $p$. Then, using the $q$ blocking sets relative to the line $\ell$, the point $p$ and a set of $q$ conics as described above, we find that there need to be $q$ more points in $A$, and hence $|A| \geq q+1$.

Case 2: no line has exactly one point in common with $A$ and there are two lines $\ell_1,\ell_2$ containing no point of $A$. Any third line $\ell$, not concurrent with $\ell_1,\ell_2$, determines a blocking system by considering $\ell \cup \ell_1 \cup \ell_2$ are removing the three intersection points. Each such line must therefore give at least two points on $A$ and hence $A$ has at least $q+1$ points.

Case 3: no line has exactly one point in common with $A$ and there is exactly one line $\ell$ containing no point of $A$. Fix a point on $\ell$; there are $q$ lines through this point that are distinct from $\ell$, and each of them contributes at least two points to $A$.

Case 4: every line has at least two points in common with $A$. This should be straightforward!

Let $\ell_1,\ell_2,\ell_3$ be three lines in the plane that do not all contain the same point. The triangle formed by $\ell_1,\ell_2,\ell_3$ is the set obtained from $\ell_1 \cup \ell_2 \cup \ell_3$ by removing the three pairwise intersections of the lines. Clearly, any triangle is a blocking set: every line in the plane meets each line $\ell_1,\ell_2,\ell_3$, and cannot contain all the three points that we are removing from the triangle. Triangles are the only blocking sets we consider.

We say that $A$ has property (*) if the intersection of $A$ with every triangle is not empty. Note that if a subset of the plane intersects non-trivially every blocking set, then it has property (*). We prove the (a priori) stronger statement that a set with property (*) and at most $q+1$ elements consists of all the points contained in a line.

Lemma 1. Let $A$ be a set of at least three points in the plane such that any line intersects $A$ in at most two points; then $A$ cannot have property (*).

Proof. Let $A' \subset A$ be a subset with three elements, and note that the points in $A'$ are not collinear. Let $\ell_1,\ell_2,\ell_3$ be the three lines each containing two of the points of $A'$. By assumption $\ell_1 \cap A , \ell_2 \cap A , \ell_3 \cap A \subset A'$; thus, the triangle formed by $\ell_1 , \ell_2 , \ell_3$ is a blocking set disjoint from $A$. $\square$

Lemma 2. Let $A$ be a set of at most $q+1$ points in the plane that are not collinear and let $\ell$ be a line in the plane such that $|A \cap \ell| \geq 3$; then $A$ cannot have property (*).

Proof. Let $a$ be a point of $\ell \setminus A$: such a point exists, since the points of $A$ are not collinear and $\ell$ contains $q+1$ points. There are $q$ lines different from $\ell$ through $a$ and at most $q-2$ points in $A \setminus \ell$; thus there are two distinct lines $\ell_1 , \ell_2$ through $a$ disjoint from $A$. Let $b$ be any point of $A$ not contained in $\ell$; there are $q+1$ lines through $b$ and only $q$ points in $A$ different from $b$; we deduce that there is a line $\ell_3$ through $b$ and disjoint from $A$. Thus the triangle formed by $\ell_1,\ell_2,\ell_3$ is a blocking set disjoint from $A$. $\square$

Corollary. Let $A$ be a set of at most $q+1$ points in the plane having property (*); then $A$ consists of all the points contained in a line.

Proof. By Lemma 1 we know that there must be a line containing three points of $A$. If the points of $A$ were not collinear, then Lemma 2 would imply that $A$ does not have property (*). Thus the points of $A$ are collinear. To conclude it suffices to show that if $A$ misses a point of the line containing $A$, then $A$ does not have property (*). This is easy: let $a$ be a point of $A$ and let $\ell_1,\ell_2$ be distinct lines through $a$; let also $b$ be a point on the line containing $A$ not in $A$, and let $\ell_3$ be any line through $b$ not containing $A$. The triangle formed by $\ell_1,\ell_2,\ell_3$ is a blocking set disjoint from $A$. $\square$