Here's what I have so far: Since $q$ is continuous and surjective then $X$ is compact. For all $x\in X$ there exists an evenly covered neighborhood $U_x$, so $\{U_x \colon x\in X\}$ is an open cover of $X$. So there is a finite subcover $\{ U_{x_1},\dots, U_{x_n}\}$.
For a contradiction, suppose $q$ is not finite-sheeted. Now for for each $x_i$, we have that $q^{-1}(U_{x_i}) = \bigsqcup_{\alpha \in I_i} V_{x_i,\alpha}$ is disjoint union of open sets of $E$. The sets $\{V_{x_1,\alpha}: \alpha \in I_1\},\dots,\{V_{x_n,\alpha}: \alpha \in I_n\}$ form an open cover of $E$, call it $\mathcal U$. So by compactness have a finite subcover $\mathcal U'\subseteq \mathcal U$. This where I'm having trouble completing the argument. Basically I want to argue that, for say $x_1$, no finite collection of $\mathcal U$ could cover the fiber $q^{-1}(x_1)$ since we assumed the fiber is infinite. But I'm having trouble how to show this.