I suppose to more formally characterize the question more formally, why are all points of the set $\{ (x,y) \mid F(x,y) = 0 \}$ always boundary points (and I believe also never isolated points) in the standard topology of $\mathbb{R}^2$ -- for simplicity's sake, say in the case where $F(x,y) \in \mathbb{R}[x,y]$ (i.e. polynomials in $x,y$ with coefficients in $\mathbb{R}$)?
I imagine there are standard proofs of such a proposition (and certainly more general than this) in algebraic geometry, but I have not studied algebraic geometry, and was wondering if there were any elementary proofs of this, only relying on basic point-set topology, algebra, real analysis, etc... I'm also looking for a relatively intuitive explanation/proof, if possible. Maybe the most intuitive reason why this is true is because of the implicit function theorem, but I wonder if there are other, still intuitive, elementary proofs of this.
Suppose $p(x,y)$ is a polynomial that is not identically $0.$ Let $Z$ be the zero set of $p.$ Then $Z$ is closed, hence $Z = \text { int } Z \cup \partial Z.$ Suppose $\text { int } Z $ is nonempty. Then there is an open disc $D(a,r) \subset Z.$ Then for any nonzero $v\in \mathbb R^2,$ the function $p_v(t) = p(a+tv)$ is a one variable polynomial in $t$ that vanishes in a neighborhood of $0.$ Thus $p_v$ vanishes identically. Since this is true for all $v,$ $p \equiv 0,$ contradiction.
This result also holds for any real analytic function $f(x,y)$ on $\mathbb R^2.$ The proof is basically the same. Beyond this, there are plenty of functions in $C^\infty(\mathbb R^2)$ for which the zero set has non-empty interior.