If $p:E \to B$ is a covering map, and if $E$ is compact, prove that $p^{-1}(b) $ is finite for all $b \in B$.
I need to verify correctness of my proof and ask if there is a more straight-forward proof for this problem.
Each $b \in B$ has an evenly covered neighborhood $U_b$, i.e. $p^{-1}(U_b)=\bigsqcup_{\alpha \in J} M_\alpha$, where $M_{\alpha}$ is an open set in $E$, which is homeopmorphic to $U_b$ under $p$, and $M_{\alpha} \cap M_{\beta} = \emptyset$ if $\alpha \neq \beta$.
$K=E \setminus (\bigsqcup_{\alpha \in J} M_\alpha)$ is a closed set in $E$, since $E$ is compact, $K$ is also compact. It is obvious that $\displaystyle\bigcup_{a \in B \ \& \ a\neq b}p^{-1}(U_a)$ is an open cover for $K$, and hence has a finite subcover $\mathcal{A}=\{A_1, \ldots,A_n\}$. Let $\mathcal{M}=\{M_{\alpha}\}_{\alpha \in J}$. Therefore $\mathcal{M} \cup \mathcal{A}$ is an open cover for $E$, hence has a finite subcover $\mathcal{A'}=\{A_1, \ldots,A_n , M_1,\ldots, M_k\}$. Now I need to show that $J=\{1,\ldots,k\}$ This has to be the case, because otherwise there exists $M_{\alpha} \in \mathcal{M}$ such that $M_{\alpha} \notin \{M_1,\ldots, M_k\}$ Obviously $M_{\alpha} \notin \{A_1,\ldots, A_n\}$ but $M_{\alpha} \in E$, contradicts the fact that $\mathcal{A'}$ covers $E$.
Your proof is certainly correct, and in full fleshed out detail there is not any way to make it more straightforward.
But one thing you can do with a proof like this is to use standard terminology to boil it down to a few pithy sentences, for example: Since $b$ has an evenly covered open neighborhood it follows that $p^{-1}(b)$ is a discrete closed subset of $E$. Every closed subset of a compact space is compact. Also, every discrete compact space is finite. It follows that $p^{-1}(b)$ is finite.
Notice what's going on here: all of your details have been moved out of your proof and into proofs of well-known, standard results about subspace topologies and compact spaces. Feel free to quote those kind of standard results.