Let $f$ be an arbitrary function from $\mathbb{R}^2$ to $\mathbb{R}$. Let $S$ be an implicit surface defined by $$ S = \lbrace x\in\mathbb{R}^2,\quad f(x) = 0 \rbrace $$
How can I prove that the implicit surface $S$ is continuous?
I feel like if $f$ is continuous, then $S$ is also continuous, but I do not know how to prove it rigorously. And I do not know if the other way is true (if $S$ is continuous, then $f$ is continuous).
I don't know how to specify a criteria about the fact that $S$ is continuous. I am trying to define two points $x,y\in S$ (so $f(x)=(y)=0$), such that $\|x - y\| < \epsilon$, $\forall\epsilon>0$. I am trying to check what happen when $x\rightarrow y$, while $x$ "stays" in $S$, but I cannot go anywhere with that.
Should I take the definition of continuity in terms of limit of function, limit of sequence, neighborhood, ...?
Should I use the gradient $\partial f$ of $f$ (which defines an orthogonal vector to the surface)? Should I impose some conditions on the gradient, like bounds, non-nullity or continuity?