TITLE: Fiber bundle is compact if base and fiber are
QUESTION [19 upvotes]: I want to show that the total space $E$ is compact if  the fiber $F$ and the base space $B$ are compact. Let $\pi$ denote the fiber projection. Since every point in $B$ has an open neighborhood $U$ whose preimage $\pi^{-1}(U)$ is homeomorphic to $U\times F$ via $\phi_U=(\pi,\beta_U)$, there are finitely many such sets covering $B$. Let's call this collection $\mathcal U$. Then $\mathcal S:=\{\phi_U^{-1}(V\times W)\mid U\in\mathcal U,V\text{ open}\subseteq U,W\text{ open}\subseteq F\}$ is a subbase for the topology on $E$.
By Alexander's subbase lemma we can restrict ourselves to considering a cover $\mathcal C\subseteq\mathcal S$ of $E$. (Mariano Suárez-Alvarez recommends this in his answer here) For each $b\in B$ the fiber $\pi^{-1}(b)$ is compact and thus contained in the union ${\bigcup_{i=1}^{n_b}\phi^{-1}_{U_i^b}(V_i^b\times W_i^b)}$ of sets in $\mathcal C$. Define $V^b:=\bigcap_{i=1}^{n_b}V_i^b$. By compactness $B=\bigcup_{k=1}^l V^{b_k}$. I think that $E=\bigcup_{k=1}^l\bigcup_{i=1}^{n_{b_k}}\phi^{-1}_{U_i^{b_k}}(V_i^{b_k}\times W_i^{b_k})$, similar to the proof of compactness of the product of two compact spaces.  
Unlike the trivial bundle case, however, where the homeo's are just identities, here the $\phi_{U}$'s could be pretty ugly, especially where they are overlapping, and we don't have $y\in\bigcup_{i=1}^{n_{b_k}}\phi^{-1}_{U_i^{b_k}}(V_i^{b_k}\times W_i^{b_k})\ $   whenever   $\ \pi(y)\in V^{b_k}$  
So I try to get to a contradiction by assuming that for each $k=1,\dots,m$ such that $\pi(y)\in V^{b_k}$ we have $\forall i=1,\dots,n_k:\ y\notin\phi^{-1}_{U_i^{b_k}}(V_i^{b_k}\times W_i^{b_k})$ which then implies $\beta_{U^k_i}(y)\notin W_i^{b_k}$. If there were a $k$ such that $U^k_1,\dots,U^k_{n_k}$ were all the same then I would be finished. But maybe the $b_k$'s should have been chosen in certain way.
Maybe someone remembers how the proof worked.

REPLY [7 votes]: I decided to answer this question because it seems that there is a plethora of proofs, none of which is the one I think of first! 

Note: I assume that compact means "compact and Hausdorff", which is the way I was taught at Warsaw University. This implies normality. 
Claim: There is a finite cover $U_{i}$ of $B$ such that $\pi$ is trivial over some open neighbourhood $V_{i}$ of the closure of $U_{i}$, ie. $\bar{U_{i}} \subseteq V_{i}$. 
Proof of the claim: By compactness it's enough to find a neigbourhood with that property of every point in $B$, so let $b \in B$. Let $V_{b}$ be an open neighbourhood of $b$. By normality, there are open sets $\{ b \} \subseteq U_{b}, (B \setminus V_{b}) \subseteq W_{b}, U_{b} \cap W_{b} = \emptyset$, because $\{ b \}, (B \setminus V_{B})$ are closed and disjoint. Now $\bar{U_{b}} \subseteq V_{b}$, since $\bar{U_{b}} \subseteq (B \setminus W_{b}) \subseteq (B \setminus (B \setminus V_{b})) \subseteq V_{b}$, so $U_{b}$ is the required neighbourhood. 

This is a basic property of (locally) compact spaces which makes precise the statement that in such spaces "neighbourhoods of points can be made arbitrarily small". The theorem you're after follows easily. 
Let $U_{i}$ be a cover of $B$ as above. Since $\bar{U_{i}} \subseteq V_{i}$, the subsets $\pi^{-1}(\bar{U_{i}}) \simeq \bar{U_{i}} \times F$ are compact and their union is $E$ by construction. So $E$ is the union of a finite number of compact subsets and thus is compact itself.