Lee in Introduction to Topological Manifolds claims that the solution set $X(f)$ of any polynomial $f\neq 0$ in two variables over $\mathbb{R}$ is nowhere dense i.e. $A:=\mathbb{R}^2-\overline{X(f)}$ is dense in $\mathbb{R}^2$. Here is what I have tried:
If for any non-empty open subset $U\subset \mathbb{R}^2$, $U\cap A=\emptyset$ then $U\cap \overline{X(f)}\neq\emptyset$. Hence, $\overline{X(f)}$ is dense in $\mathbb{R}^2$, i.e. $\overline{X(f)}=\mathbb{R}^2$.
At this point I am stuck or I am not sure whether what I am doing is right: judging by the fact that Zariski topology (I am also doing some commuatative algebra) on $\mathbb{R}^2$ is coarser than the Euclidean topology, the algebraic set $X(f)$ is closed with respect to the Euclidean topology and hence $X(f)=\overline{X(f)}=\mathbb{R}^2$. I believe that this is true iff $f$ is the zero polynomial, hence I have a contradiction.
Here I don't really need the Zariski topology to give me the closedness. Instead, thanks to a friend, it would be simply enough to note that the solution set is pre-image $f^{-1}(0)$ of $0$ which is a closed set in $\mathbb{R}^2$. Finally, primages of closed sets under continuous maps are closed. But I believe the proof is structurally wrong, for now.
I would greatly appreciate if any mistakes are pointed out. Also I would be thankful for alternative ways to approach this problem.