TITLE: Finite covering is compact Hausdorff iff base space is
QUESTION [8 upvotes]: I am in need of solution or tip for this question. I thank you.
Let $ p: \widetilde X \to X $ be a covering space with $ p^{-1}(x) $ finite and nonempty for all $ x \in X$. Show that $ \widetilde X$ is compact Hausdorff iff $ X$ is compact Hausdorff.

REPLY [6 votes]: As you had got hint from the previous answer, hope you have completed the solution. Here I want to give the idea of the proof of this statement from another point of view which does not require an open covering of $ \widetilde{X} $.(I'll show the difficult part only)
All we have to note is that a compact Hausdorff space is regular. So the base space   $\ X $ is regular. Now first cover $\ X $ with evenly covered neighbourhoods 
{$\ U_{\alpha} $} . Using regularity condition we can get a refinement of this open covering - say {$\ V_{\beta} $} - such that closure of $\ V_{\beta} $ is inside some $\ U_{\alpha} $. 
So, $\ p^{-1} (\overline{V_{\beta}})$ is disjoint union of finitely many closed sets. (Using the fact fibres are finite and $ U_{\alpha} $  is an evenly covered neighbourhood).  Now $\ \overline{V_{\beta}} $ is compact so is its preimage because the covering map is local homeomorphism. Now get a finite subcover of $\ X$ from the collection {$\ V_{\beta} $} and hence preimage of the corresponding $\ \overline{V_{\beta}} $ 's will cover $ \widetilde{X} $ . So we get $ \widetilde{X} $ is union of finitely many compact sets which implies  $ \widetilde{X} $ is compact.