Fiber bundle map is proper if the model fiber is compact This is one direction of problem 10-19 (c) from John Lee's Introduction to Smooth Manifolds.
Suppose $\pi: E \to M$ is a fiber bundle with fiber $F$.
Show that $\pi$ is a proper map if $F$ is compact.
Here, $M$, $E,F$ are topological spaces with no further assumption and $\pi:E \to F$ is a surjective continuous map with the property that for each $x\in M$, there exist a neighborhood $U$ of $x$ in $M$ and a homeomorphism $\Phi: \pi^{-1}(U)\to U\times F$, called a local trivialization of $E$ over $U$, such that we have $\pi_1 \circ \Phi = \pi$, where $\pi_1$ is the projection onto $U$.
I have seen a solution to this problem Let $\pi: E \rightarrow M$ be a fiber bundle with fiber $F$. Then $\pi$ is a proper map $\iff F$ is compact.
here that uses the manifold assumptions on $E$ and $M$, namely Hausdorff and local compactness, but I cannot figure out a way to prove this in the general case without these assumptions on the topological spaces.
Suppose that $F$ is compact and $A \subset M$ is compact. Then we need to show that $\pi^{-1}(A)$ is compact. Clearly we would need to use the local trivialization property, so that for each $p \in A$, we can find a cover $U_p$ such that $\Phi: \pi^{-1}(U_p) \to U_p \times F$ is a homeomorphism and $p \in U_p$. And cover $A$ with finitely many $U_i$'s. But I cannot think of a way to progress from here without resorting to local compactness.
How can we prove this? I would greatly appreciate some help.
 A: Let $V_i$ be an open cover of $\pi^{-1}(A)$, with $A$ compact, and pick sets
$$V^x_i$$
from $\{V_i\}$ such that the collections $\{V_i^x\}$ are finite subcovers of each fiber $F_x$. Then consider
$$U_x = \left\{y \in M | \{V_i^x \} \text{ covers } F_y \right\}.$$
These sets are non empty since $x \in U_x$ and open. If
$$
\phi : \pi^{-1}(X) \to X \times F
$$
is a trivialization, then
$$
U_x \cap X = X \setminus \pi\left(\pi^{-1}(\bar X) \setminus \bigcup V_i^x\right).
$$
The left hand side is open and hence for any $p \in U_x$, picking $X$ such that $p \in X$ gives that $U_x$ is a neighborhood of $p$. Hence $\{U_x\}$ is an open cover of $A$ so we can pick $x_1, \dots, x_n$ such that
$$
A \subseteq \bigcup_{j = 1}^nU_{x_j}.
$$
Let $B = \{x_1, \dots, x_n\}. $Then the collection $$\{V_i^{x} | x \in B, i \in \{1, \dots , n_x\} \}$$ covers $\pi^{-1}(A)$.
A: Here is a proof using nets. Recall that a space $X$ is compact if and only if every net in $X$ contains a convergent subnet (see e.g. this question). Let $K\subset M$ be a compact, $C:= \pi^{-1}(K)$. Let $x_\bullet$ be a net in $C$. Then, by compactness of $K$, the net $\pi(x_\bullet)$ contains a subnet converging to a point $k\in K$. This  subnet has the form $\pi(y_\bullet)$, where $y_\bullet$ is a subnet in $x_\bullet$. Since $\pi(y_\bullet)\to k$, WLOG we can assume that $\pi(y_\bullet)$ is contained in a neighborhood $U$ of $k$ (in $M$) over which the bundle is trivial. By trivializing the bundle over $U$, we obtain that $y_\bullet$ is contained in $U\times F$. Let $\pi': U\times F\to F$ be the projection to the second factor. Since $F$ is compact, the net $y_\bullet$ contains a further subnet $z_\bullet$ such that $\pi'(z_\bullet)$ converges to some $f\in F$. Since $\pi(z_\bullet)$ still converges to $k\in K$, we obtain that $z_\bullet$ converges to $(k,f)\in C=\pi^{-1}(K)$. Since $z_\bullet$ is a subnet in $x_\bullet$ (a subnet of a subnet is again a subnet), we obtain that the net $x_\bullet$ contains a subnet converging to a point in $C$. Since $x_\bullet$ was arbitrary, $C$ is compact.
Lastly, I am not sure that the definition of properness in terms of compacts is useful once you work with non-Hausdorff spaces. Algebraic geometers (working with Zariski topology), tend to use a different notion, the one of universally closed maps.
A: If my memories are correct, then the fact that the fiber is compact implies that the projection map $\pi$ is a closed map. Now it is theorem of general topology that a set $K$ is compact if and only if $K$ satisfies the following property: If $\{F_\alpha\}_{\alpha\in A}$ is an arbitrary family of closed sets in $K$ such that any finite number of $F_\alpha$ has non-trivial intersection, then $\cap_{\alpha \in A} F_\alpha\neq \emptyset $. We use this characterisation of compactness.
Now suppose you have $K\subseteq M$ compact, and you have a family of closed sets $\{F_\alpha\}_{\alpha\in A}$ in $\pi^{-1}(K)$, such that any finite number of $F_\alpha$ has non-trivial intersection. Let $\widetilde{A}=\{ \text{finite subsets of } A \}$, if $I\in \widetilde{A}$, then let $G_I=\pi(\cap_{i\in I}F_i)$. Notice that $\cap_{i\in I}F_i\neq \emptyset$ since $I$ is a finite subset of $A$, thus $G_I\neq \emptyset$. Since finite intersection of closed sets is closed and $\pi$ is a closed map we also see that $\{G_I\}_{I\in \widetilde{A}}$ are closed sets. Now if $I_1,\dots,I_N\in \widetilde{A}$, then I claim that $\cap_{j=1}^N G_{I_j}\neq \emptyset$. Indeed $\cap_{j=1}^N G_{I_j}=\cap_{j=1}^{N}{\pi(\cap_{i\in I_j}F_i)}\supseteq \pi(\cap_{j=1}^N(\cap_{i\in I_j}F_i))\neq \emptyset$. In the last step we use that $\cap_{j=1}^N(\cap_{i\in I_j}F_i)\neq \emptyset$ since it is a finite intersection. Now, since any finite intersection of $\{G_I\}_{I\in \widetilde{A}}$ is non-empty and $K$ is compact we get that $\cap_{I\in \widetilde{A}}G_I\neq \emptyset$. Let $p$ be in the intersection, and let $H_\alpha=\pi^{-1}(p)\cap F_\alpha$. Now $H_\alpha\neq \emptyset$ since $\pi(F_\alpha)=G_I$, with $I=\{\alpha\}$, and clearly it is a closed subset of $\pi^{-1}(p)$. Let $\alpha_1,\dots,\alpha_N\in A$, then $\cap_{i=1}^N H_{\alpha_i}=\pi^{-1}(p)\bigcap\cap_{i=1}^N F_{\alpha_i}\neq \emptyset$, since for $I=\{\alpha_1,\dots,\alpha_N\}$ we see that $p\cap \pi(\cap_{i=1}^N F_{\alpha_i})=p\cap G_I=\{p\}\neq \emptyset$. Thus $\{H_\alpha\}_{\alpha\in A}$ also satisfies that any finite intersection is non-empty, and since $F$ is compact we get that there exists an element $q$ in $\cap_{\alpha\in A}{H_\alpha}\subseteq \cap_{\alpha\in A}F_\alpha$, thus $\pi^{-1}(K)$ is compact.
