Compactness and Covering Spaces Let p : Y -> X be an n-sheeted covering map, where X and Y are topological spaces. If X is compact, prove that Y is compact.
I realize that this seems like a very simple problem, but I want to stress the lack of assumptions on X and Y. For example, this is very easy to prove if we can assume that X and Y are metrizable, for sequential compactness is then equivalent to compactness and it is easy to lift sequential compactness from X to Y.
I asked three people in person this question and all of them immediately made the assumption that X and Y are metrizable, so I feel like I should put in this warning here that they are not.
 A: Well, the obvious argument that any sequence has a convergent subsequence that your three friends used for the metrizeable case generalizes easily to show that any net has a convergent subnet in the general case.
A: A direct argument without the use of nets:
Let $\mathcal{C}$ be an open cover of $Y$.  For each $p \in X$, choose an open set $p \in U \subseteq X$ such that $Y$ is trivial over $U$, and such that each lift of $U$ is contained in some element of $\mathcal{C}$.  This is an open cover $\mathcal{D}$ of $X$, which has a finite subcover $\mathcal{D}'$ since $X$ is compact.  The lift of $\mathcal{D}'$ to $Y$ is also a finite cover, as well as a cover that refines $\mathcal{C}$.  Thus $\mathcal{C}$ must have a finite subcover.  (The fact that $Y$ is a finite cover is used twice, first to make each $U$, second to lift $\mathcal{D}'$.)
A: Dear Eric, here is a Bourbaki-style proof.
Recall that a continuous map $f: Y\to X$ is called proper by Bourbaki if, for all spaces $Z$, the map  $f\times 1_Z: Y \times  Z\to X \times Z$ is closed.  For example the trivial finite covering 
$X\times \{ 1,\ldots n \}\to X$ is proper.
Now, your $X$ is covered by opens $X_\iota \subset X$ such that the restricted/corestricted maps $f_{X_\iota }:f^{-1} (X_\iota) \to X_\iota $ are trivial finite coverings, hence are proper by the example above. We deduce that the original covering $f:Y\to X$ is proper: this follows easily from the definition of "proper" and (if a reference is needed) is proved in Bourbaki's General Topology, Chapter 1, §10, Proposition 3.
But a proper map has the property that the inverse image of a quasi-compact subset of the target (in our case all of $X$) is quasi-compact (ibid., Proposition 6). Hence $Y$ is quasi-compact if $X$ is.
NB I have used  Bourbaki's definition "universally closed" for proper. As I said, this implies  that inverse images of quasi-compact subsets are quasi-compact.This last property is often taken as the definition of proper. For locally compact spaces, both definitions coincide.  
