Given a fiber bundle $F\to E\overset{\pi}{\to} B$ such that $F,B$ are compact, is $E$ necessarily compact?

Consider a (locally trivial) fiber bundle $F\to E\overset{\pi}{\to} B$, where $F$ is the fiber, $E$ the total space and $B$ the base space. If $F$ and $B$ are compact, must $E$ be compact?

This certainly holds if the bundle is trivial (i.e. $E\cong B\times F$), as a consequence of Tychonoff's theorem. It also holds in all the cases I can think of, such as where $E$ is the Möbius strip, Klein bottle, a covering space and in the more complicated case of $O(n)\to O(n+1)\to \mathbb S^n$ which prompted me to consider this question. I am fairly certain it holds in the somewhat more general case where $F,B$ are closed manifolds. However, I can't seem to find a proof of the general statement. My chief difficulty lies in gluing together the local homeomorphisms to transfer finite covers of $B\times F$ to $E$. Any insight would be appreciated.

By local triviality, there is a open covering $\mathcal U$ of $B$ such that for each $U\in\mathcal U$ the open subset $\pi^{-1}(U)$ of $E$ is homeomorphic to $U\times F$ in a way compatible with the projection to $U$. It follows that there is a subbase $\mathcal S$ of the topology of $E$ consisting of open sets each of which is contained in one of these $\pi^{-1}(U)$ and corresponding under those homeomorhisms to an open subset of $U\times F$ of the form $V\times W$ with $V\subseteq U$ open in $B$ and $W\subseteq F$ open in $F$.

To prove compactness, it is enough to show that every covering of $E$ by subsets of $\mathcal S$ contains a finite subcovering —this is called Alexander's subbase lemma and is used in one of the proofs of Thychonof's theorem (for example, in Kelley's book, iirc). Do that!

I don't get where the problem is. Am I missing something?

Each point $b\in B$ has a neighbourhood $N_b$ such that the bundle over $N_b$ is trivial. Choose a smaller closed (thus compact) neighbourhood $C_b$ (we need some weak assumption here like Hausdorffness of $B$). The bundle over $C_b$ is homeomorphic to $C_b\times F$, thus compact.

Note that $\{\mathrm{int}(C_b)\}_{b\in B}$ is an open cover of $B$ and by compactness has a finite subcover indexed by $(b_i)_{i=1}^n$. Consequently $E$ is a finite sum of compact sets, thus compact: $$E = \bigcup_{1\leq i \leq n} \pi^{-1}(C_{b_i}).$$