Let $p: E\to B$ be a covering map. If $B$ is compact and $p^{-1}(b)$ is finite, then $E$ is compact. So I start off and assume that some $\{U_\alpha\}$ is a cover of $E$. 
I want to reduce this cover to a finite subcover of $E$.  
Since $p$ is a covering map it is also an open map, therefore $p(\{U_\alpha\})$ is an open cover of $B$ denote it $\{W_\alpha\}$. 
But since $B$ is compact there is a finite subcover of $\{W_\alpha\}$, $\cup_{i=1}^nW_i$.
But then since $p$ is continuous $p^{-1}(\cup_{i=1}^nW_i)$ is an open cover of $E$ which must stay finite since $p^{-1}(b)$ is finite for all $b$.
 A: You could use the fact that a covering map with finitefibers is a perfect map, i.e. a surjective and closed map with compact fibers. Closedness of $p$ is shown as follows:
Let $C\subset E$ be closed, $b\notin p[C]$. Then $p^{-1}(b)=\{b_1,...,b_n\}$ is a finite subset of $E-C$. There is an evenly covered neighborhood $U$ of $b$ whose preimage is the union of $U_i$, $i=1,....,n$ so that $b_i\in U_i$. Now each $b_i$ has a neighborhood $V_i$ contained in $U_i\cap(E-C)$. If $x\in \bigcap_i p[V_i]$, then its fiber has one point in each $U_i$ which must thus be in $V_i$, so $x$ cannot be in $p[C]$.
Now perfect maps are proper, meaning that preimages of compact sets are compact.
Also see my answer for the more general case of a fiber bundle with compact fiber $F$. A covering space can be characterized as a fiber bundle where the fiber is discrete.

Instead of looking at the images of the $U_\alpha$, consider the following sets
$$U^b=\text{ a finite subcover of $p^{-1}(b)$ from the $U_\alpha$}\\
W_b=p_*(U^b)=B-f(E-U^b)=\{c\in B\mid f^{-1}(c)\subseteq U^b\}$$
This $p_*$ is also called the dual image and it is inclusion preserving. It's easy to see that $p^{-1}(W_b)\subseteq U^b$. Furthermore, since $p$ is closed, then $p_*(U^b)$ is an open neighborhood of $b$, so the $W_b,\ b\in B$, form an open cover of $B$. There are now $b_1,...,b_n$, so that $W_{b_i}$ cover $B$. It follows that the $U^{b_i}$ cover $E$.
