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!