TITLE: Compactness about fiber bundle
QUESTION [6 upvotes]: I am working on a problem (problem 10.19 (d)) in John M. Lee's Introduction to Smooth Manifold.
Assume that $\pi$: $E$ $\rightarrow$ $M$ is a fiber bundle with model fiber $F$, I need to prove if $E$ is compact, then so are $M$ and $F$.
Clearly, $M$ is compact and if we assume that $M$ is Hausdorff, then it follows easily from part (c) of this problem.
($\pi$: $E$ $\rightarrow$ $M$ is proper iff $F$ is compact.)
(Edited: I think it suffices to assume that $M$ is $T_1$, i.e., singletons are closed, then it follows from the continuity that the fibers are compact.)
However, I have no idea how to proceed in the general case.
Any hints are appreciated.

REPLY [6 votes]: Since the author chimed in in the comments, let me record the full answer here.
This is still true without separation assumptions on the base. Indeed, by compactness of $M$ there exists a finite cover $\{U_i \subseteq M\}_{i \in I}$ over which $E$ is trivialised. Without loss of generality we may assume that no strict subset of $\{U_i \subseteq M\}$ covers $M$. Then
$$Z = M \setminus \bigcup_{j \neq i} U_j \subseteq U_i$$
is a nonempty closed subset over which $E$ is trivial, i.e. $E|_Z \cong F \times Z$. Since $E|_Z \subseteq E$ is closed, it is compact, hence so is $F$ since it is the image of the first projection $F \times Z \to F$. $\square$