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.
Best Answer
We use the following characterization of compactness: every infinite net has a convergent subnet.
Let $\{x_{\lambda}\}_{\Lambda}$ be an infinite net in $E$, $\Lambda$ the directed set of the net. As $B$ is compact, there is an infinite subnet $\Lambda_1 \subset \Lambda$ for which $\{\pi(x_{\lambda})\}_{\Lambda_1}$ converges to a point $b \in B$. Fix a local trivialization $\phi_U : \pi^{-1}(U) \rightarrow U \times F$ for $b \in U$. Pare off the net $\Lambda_1$ so that $\pi(x_{\lambda}) \in U$ for all $\lambda$.
Now let $\pi_2 : U \times F \rightarrow F$ be the continuous projection onto the second factor. So, $\{\pi_2 \circ \phi_U(x_{\lambda})\}$ is an infinite net in $F$, hence possesses yet another convergent subnet $\Lambda_2$. Then $\{\phi_U(x_{\lambda})\}_{\Lambda_2}$ is convergent in $U \times F$ by definition of the product topology, and the image of this convergent net under the homeomorphism $\phi_U^{-1}$ is convergent. Thus $\{x_{\lambda}\}_{\Lambda_2}$ is a convergent subnet.