Difficulty: 2

Let $X$ be infinite and $A \subseteq X$ countably infinite (uses axiom of countable choice). If $X$ has no limit points, $A$ is closed and for each $x \in A$, choose a nbd $V_x$ with $V_x \cap X = \emptyset$. Then $\{V_n\} \cup \{A^C\}$ is a cover. A finite subcover would imply $A$ is finite.

For $f : X \to \R$ continuous, $\{f^{-1}(-n, n)\}$ is a countable cover.

It has a bijection $f: X \to n$ if finite, or $f : X \to \omega$ if infinite. It’s a homeomorphism either way.