Let $\pi: E \rightarrow M$ be a fiber bundle with fiber $F$. Then $\pi$ is a proper map $\iff F$ is compact.
I'm assuming everything is smooth. The "$\implies$" direction is trivial but I'm struggling with the converse. I have already proved that $\pi$ is an open quotient map and a smooth submersion in a previous part of the problem. I thought maybe the fact that $\pi$ is an open map would be useful.
I take a compact subset $K \subset M$ and look at an open cover $\{U_{\alpha}\}_{\alpha \in A}$ and want to try to find a finite subcovering of that. I've tried looking at the images $V_{\alpha}=\pi(U_\alpha)$ which are an open cover of $K$ and reducing to a finite subcover by compactness, but I can't figure out how to make this work. Obviously we need to use the assumptions on $F$ being compact so we probably need to bring in local trivializations somewhere.
Best Answer
Consider the family $\mathcal F$ of the open subsets $U\subseteq M$ such that:
$\overline U$ is homeomorphic to the closed unit ball;
there is some trivializing open set $V$ such that $V\supseteq \overline U$.
$\mathcal F$ is a basis for the topology of $M$. Therefore, there is a finite cover of $K$ by elements of $\mathcal F$. Now, $\pi^{-1}[K]$ is a closed subset of $\bigcup_{j=1}^m \pi^{-1}\left[\overline U_j\right]$. By definition there are homeomorphisms $\chi_j$ defined on neighbourhoods of $\pi^{-1}\left[\overline U_j\right]$ that send $\pi^{-1}\left[\overline U_j\right]$ to the compact topological space $\overline U_j\times F$. Therefore, $\pi^{-1}[K]$ is a closed subset of a finite union of compact sets, and thus compact.