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Prove that a Covering map is proper if and only if it is finite-sheeted.

First suppose the covering map $q:E\to X$ is proper, i.e. the preimage of any compact subset of $X$ is again compact. Let $y\in X$ be any point, and let $V$ be an evenly covered nbhd of $y$. Then since $q$ is proper, and $\{y\}$ is compact, $q^{-1}( \{ y\})$ is also compact. In particular the sheets $\bigsqcup_{\alpha\in I}U_\alpha$ of V are an open cover of $q^{-1}( \{ y\})$ and must therefore contain a finite subcover $\{U_1,...,U_n\}$. Then the cardinality of the fiber $q^{-1}( \{ y\})$ is $n$, so that $q$ is finite-sheeted.

Conversely, we suppose that $q$ is finite-sheeted. Let $C\subset X$ be a compact set, and let $\{U_a\}_{a\in I}$ be an open cover of $q^{-1}(C)$...

Now how do I continue?

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$\Rightarrow$ Each fiber is compact (by properness) and discrete (from definition of covering space) hence is finite.

$\Leftarrow$ You have to prove that for $K\subset X$ the inverse image $q^{-1}(K)$ is compact.
Since $\operatorname {res} q:q^{-1}(K) \to K$ is a finite covering space in its own right apply my answer to cocomi.

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"$\Rightarrow$" Since a singleton $\{x\}\subset X$ is compact wrt every topology, $q^{-1}(\{x\})$ is compact. Covering gives the property for $q^{-1}(\{x\})$ to be discrete in the sense that there exists an open neighborhood $V \ni x$ such that each connected component of $q^{-1}(V)$ is homeomorphic to $V$ itself. Then clearly the connected components are disjoint open sets. This means that for each $e\in q^{-1}(\{x\})$ there exists an open neighborhood $U\ni x$ such that $U\cap q^{-1}(\{x\}) = \{e\}$ (definition of discrete set). Compactness implies sequential compactness therefore if $q^{-1}(\{x\})$ wasn't finite we could extract a sequence which does not converge to any of its points (because they are very separated by open sets). Absurdness which clearly shows that $q^{-1}(\{x\})$ is finite.

"$\Leftarrow$" Let $K\subset X$ be compact and $L=q^{-1}(K)$. If $K$ is was empty than the result would be trivial so assume $K\neq \emptyset$. Let $\{T_\alpha\}$ be an open covering for $L$. Openness of $q$ implies that $\{q(T_\alpha)\}$ is an open covering for K. For each $x\in K$ there exists $U_x \subset q(T_{\alpha})$ (for some $\alpha$) open neighborhood such that each connected component of $q^{-1}(U_x)$ is homeomorphic to $U_x$. Assume now that the components of $q^{-1}(U_x)$ are not finite. Then there are components $E_1,\ldots,E_n \subset q^{-1}(U_x)$ containing $\{e_1,\ldots,e_n\} = q^{-1}(\{x\})$. The others cannot contain them and therefore they are not bijective onto $U_x$ since they don't map anything into $x$ itself. Since these are finite we can shrink them in a way that if a component of $q^{-1}(U_x)$ has non empty intersection with $L$, than it is a subset of $T_\alpha$ for some $\alpha$. Cover $K$ by finitely many such sets $K\subset \cup_{i=1}^m U_{x_i}$. Now $L\subset q^{-1}(\cup_{i=1}^m U_{x_i}) = \cup_{i=1}^m q^{-1}(U_{x_i})$ is covered by finitely connected components and each of these is inside a $T_\alpha$, therefore $L$ is covered by finitely many $T_\alpha$, i.e. it is compact.

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