Covering map is proper $\iff$ it is finite-sheeted 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?

$\Rightarrow$ 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.