Construct a set $A\subset \mathbb{R}^2$ with $|A|=6$ such that for all $f\in \mathbb{R}[x,y]$ with $\deg(f)\leq 2$ then $|A\cap V(f)|\leq 5$ Where $V(I)$ denote the set of zeros of $I\subset K[x_1,x_2,\cdots ,x_n]$.
I don´t know how I should construct a set, for instance I think that it follows of apply Schwartz-Zippel Lemma, but unfortunally $A\subset \mathbb{R}^2$ can´t be divided in sets $A_1,A_2,A_3,\cdots A_n$ such that for $f\in \mathbb{R}[x,y]$ we should have $|V(f)\cap (A_1\times A_2\times \cdots \times A_n)|\leq dm^{n-1}$.
Other idea less mathematically is think that if $f$ have degree less or equal to $2$ then $f$ must be of the form $$f(x,y)=ax^2+bx+c$$, $$f(x,y)=ay^2+by+c$$, $$f(x,y)=ax+by+c$$, $$f(x,y)=axy+by+cx+d$$ $$f(x,y)=Ax^2+By^2+Cx+Dy+Exy+F$$
And then consider $A=\lbrace a_i=(x_i,y_i)\mid x_i,y_i\in \mathbb{R}\rbrace$ and $i\in\lbrace 1,2,3,4,5,6 \rbrace$ and then make $f(a_i)=0$ and show that the sistem of equations have no trival solutions for any chose of $A,B,C,D,E,F$
The statement clearly fails when $f$ is the zero polynomial. I'll assume you want an $A$ which satisfies $|A \cap V(f)| \le 5$ for all non-zero polynomial with degree at most $2$.
For any $p = (x,y) \in \mathbb{R}^2$, let $\vec{p} =(x^2,xy,y^2,x,y,1) \in \mathbb{R}^6$.
For any $f \in \mathbb{R}[x,y]$ with $\deg f \le 2$, let $B, \ldots, G$ be its coefficients. More precisely, $$f(x,y) = B x^2+ Cxy+Dy^2+Ex+Fy+G$$ Let $\vec{f} = (B,C,D,E,F,G) \in \mathbb{R}^6$.
It is easy to see $p \in V(f)$ if and only if $\vec{f}\cdot \vec{p} = 0$.
If you pick any $6$ points $p_1,\ldots,p_6$ so that corresponding $\vec{p}_1,\ldots,\vec{p}_6$ are linear independent, then $\vec{f}\cdot \vec{p}_i = 0$ for all $i$ forces $\vec{f} = \vec{0}$ which is equivalent to $f$ is the zero polynomial.
To see this, treat $\vec{f}$ and $\vec{p}_i$ as $1\times 6$ row vectors. Let $P$ be the $6\times 6$ matrix whose $i^{th}$ row is $\vec{p}_i$. It is not hard to see
$$\vec{f} P^T = (\vec{f}\cdot\vec{p}_1,\ldots,\vec{f}\cdot\vec{p}_6)$$ The condition $\vec{f}\cdot\vec{p_i} = 0$ for all $i$ is equivalent to $\vec{f} P^T = \vec{0}$. If $\vec{p}_i$ is linear independent, $P$ and hence $P^T$ will be invertible and this leads to $\vec{f} = \vec{0} (P^T)^{-1} = \vec{0}$.
For such $6$ points, let $A = \{ p_1, \ldots, p_6 \}$. It will have the property $|A \cap V(f)| \le 5$ for all non-zero polynomial $f$ with degree at most $2$. As an example, you can choose $$(p_1,\ldots,p_6) = (0,0),(1,0),(0,1),(-1,0),(0,-1),(1,1)$$ By brute force, you can verify $\det P = 4$, this implies above $\vec{p}_1,\ldots,\vec{p}_6$ are linear independent and give you one example of $A$.