Let $ X $ be compact and let $f: X \rightarrow Y$ be a local homeomorphism. Show that for any point $y \in Y$, $f^{-1}(y)$ is a finite set.

Without going into evenly covered maps etc., it might be simpler: Suppose for some $y$, $F_y := f^{-1}[\{y\}]$ is infinite. As $X$ is compact (in fact we need only countably compact) there is a limit point $p$ of $F_y$. But if $U$ is a neighbourhood of $p$ such that $f \restriction U: U \to f[U]$ is a homeomorphism, then $U$ contains at least 2 (in fact infinitely many; here I'm assuming $X$ is $T_1$ as well) points of $F_y$, which makes $f$ not even injective on $U$, as both map to $y$ under $f$.

So in fact, for a locally injective map $f$ on a countably compact $T_1$ space, all fibres are finite.

So if some $F_y$ were infinite, by compactness it would have an $\omega$-accumulation point $x \in X$. If $O$ were the witnessing neighbourhood of $x$ wrt local homeomorphism, a contradiction to injectivity ensues; no sep. axioms needed.

Let me suggest a strategy. The problem is to show that every fiber $f^{-1}(y)$ is finite when $f$ is a local homeomorphism; if it's unclear what the role of this hypothesis is, try asking the same question without it: is it necessarily true that the fiber of any map $f$ is finite, when $X$ is compact? Obviously not: you could have, as a really silly example, the map $f \colon X \to \{\text{pt}\}$, whose fiber is all of $X$, which doesn't need to be finite. Though if such an $f$ were a local homeomorphism, then $X$ would be a discrete union of points and a discrete, compact space is finite. So the claim holds for maps from a compact space to a point.

It's crucial to recall the proof of the last claim there. If $D$ is a discrete compact space, then each point is, in itself, an open set (discreteness), and $D$ is the union of its points, so is a finite union of some of them (compactness). But since each singleton set contains only one point, this accounts for only finitely many points, so $D$ is finite.

So if we want to execute the same argument on a general map $f \colon X \to Y$, we need to cover the fiber $f^{-1}(y)$ by open sets each of which contains only one point of the fiber, and somehow argue by compactness that there is a finite subcover. Note that (without some kind of separation axiom) the fiber itself may not be compact, but this is not an impediment as one can always enlarge the covering by adding irrelevant open sets.

All the ingredients for this are in your outlined proof.