Let $P(x,y)=\sum_{m,n=0}^{1000}a_{mn}x^{m}y^{n} $ with $ a_{mn}\in \mathbb{R}$.
Suppose that the zero set $\left \{ (x,y)\in \mathbb{R}^2 : P(x,y)=0) \right \}$ contains an open neighborhood in $\mathbb{R}^2$.
Then prove that $P(x,y)\equiv 0 $ on $\mathbb{R}^2$.
I thought that this is similar to Identity theorem. Identity theorem: If $P(x)$ and $Q(x)$ are polynomials and $P(x)\equiv Q(x)$ on a small interval, then $P(x)\equiv Q(x)$ in $\mathbb{R}$. But, I don't know how to proceed.
Without loss, we can suppose that $P(x,y)=0$ on a ball centered at $0$ just by shifting. Now, for $k\in\mathbb{R}$, consider $p_k(x):=P(x,kx)$. This is now a polynomial on $\mathbb{R}$ and is zero on an open interval, so, as you pointed out, we have $p_k(x)\equiv 0$ for all $k$. Thus, for all $x\neq 0$, we have $P(x,y)=0$ for all $y\in\mathbb{R}$. This means that $P\equiv 0$ since $\mathbb{R}^2\setminus\{(0,y):y\in\mathbb{R}\}$ is dense in $\mathbb{R}^2$. (Or you can do a symmetric argument using $P(ky,y)$).