Theorem 1   Let $C$ be a countable subset of $\mathbb{R}^2$ . Then $\mathbb{R}^2\setminus C$ is path connected.

Proof. Fix a point $P$ not in $C$ . The strategy of the proof is to construct a path $p_x$ from any $x \in \mathbb{R}^2\setminus C$ to $P$ . If we can do this then for any $d, d' \in \mathbb{R}^2\setminus C$ we may construct a path from $d$ to $d'$ by first following $p_d$ and then following $p_{d'}$ in reverse.

Fix $x \in \mathbb{R}^2\setminus C$ , and consider the set of all (straight) lines through $x$ . There are uncountably many of these and they meet in the single point $x$ , so not all of them contain a point of $C$ . Choose one that doesn't and move along it: your distance from $P$ takes on uncountably many values, and hence at some point this distance $r$ from $P$ is not shared by any point of $C$ . The whole of the circle with radius $r$ , centre $P$ , lies in $\mathbb{R}^2\setminus C$ so we may move around it freely.

Consider all lines through $P$ : these all intersect this circle, and there are uncountably many of them so we may choose one, say $L$ , that contains no point of $C$ . Moving around the circle until we meet $L$ and then following it inwards completes our path form $x$ to $P$ .

Corollary 1   Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be continuous and onto. Then $f^{-1} (0)$ is uncountable.

Proof. Suppose that $f^{-1} (0)$ is countable. $\mathbb{R}^2$ can be written as the disjoint union $$ f^{-1} (0) \cup f^{-1} ((-\infty, 0)) \cup f^{-1} ((0, \infty)) $$ where the last two sets are open (as $f$ is continuous), non-empty (as $f$ is onto) and disjoint. Since pathwise connected is the same as connected for Hausdorff spaces, we have that $\mathbb{R}^2 \setminus f^{-1} (0)$ is not path connected, contradicting the theorem.