Problem statement: Show that $R^2$ cannot be written as a countable union of zero sets of non-trivial polynomials. Note that the zero set of a polynomial $p(x,y)$ is $\{(x,y) : p(x,y) = 0$}.
My attempt: I'm fairly sure that this is a Baire category theorem problem. If we could show that any zero set of a non-trivial polynomial is nowhere dense, ie, has an empty interior, we would be done, since $R^2$ is locally compact, I think. However, I'm unsure of how to show this.
So far, I have written that, if $A = \{(x,y) : p(x,y) = 0\}$, for some polynomial $p$, and $(x',y') \in \left(\bar{A}\right)^o$, then for some $\epsilon > 0$, there exists $B((x',y'),\epsilon) \subset \bar{A}$. Now, we also have that if $(a,b) \in B((x',y'),\epsilon)$, for any $r$, there exists some point $(\alpha,\beta) \in B((a,b),r)$ such that $p(\alpha,\beta) = 0$. Theoretically, there should be a contradiction somewhere here, but I can't see it.
Another idea to mess around with was to possibly try showing that each zero set has measure $0$, and therefore $R^2$ could not be written as the countable union of $0$ sets... I immediately jumped to trying to use Baire Category, but it's entirely possible that I'm just doing this all wrong!
Basically, I'm stuck here, and I would really appreciate a hint in the right direction on where to go from here.
Hint: if $f \in \mathbb{R}[t]$, $f$ has either finitely many roots, or $f$ is identically zero. So if $p \in \mathbb{R}[x, y]$ and $p$ vanishes on an open subset $U$ of $\mathbb{R}^2$, then $p$ vanishes on $\mathbb{R}^2$, because, for any $(x_0, y_0) \in U$ and every $(x, y) \in \mathbb{R}^2$, the polynomial $q(t) = p(x_0 + tx, y_0 + ty)$ has infinitely many roots and hence is identically zero. So the zero set of a non-trivial $p$ is nowhere dense and you can apply the Baire category theorem.