A proper local diffeomorphism between manifolds is a covering map.

Solution 1:

In the case of manifolds, you can use the characterization of covering map via the path-lifting property:

$f$ is a covering if for every $x\in M$, for every continuous path $\sigma:[0,1]\to M$ with $\sigma(0)=x$, and for every $\tilde x\in\tilde M$ that prjects to $x$, there is a unique lift $\tilde\sigma:[0,1]\to \tilde M$ that projects to $\sigma$ (i.e. $f(\tilde\sigma(t))=\sigma(t)\ \forall t$).

Now, the properness of $f$ plus the local diffeo porperty ensure the path-lifting property:

Let $x,\tilde,\sigma$ be as above. Let $D\subset[0,1]$ the set of times for which $\tilde\sigma$ exists: $D=\{t\in[0,1]: \tilde\sigma$ is defined on $[0,s]$ for every $s\leq t\}$.

We show that $D=[0,1]$ by showing that it is open and closed and non-empty. Clearly $D\neq\emptyset$ because it contains $0$ by definition of $\tilde x$. The local diffeo property of $f$ implies that $D$ is open.

To see that $D$ is closed we use properness. Let $T=\sup D$ so that $\bar D=[0,T]$ is compact. By properness $f^{-1}([0,1])$ is compact. Thus its connected components are compacts. Let $S$ be the connected component containing $\tilde x$. $S$ is compact hence is closed. Since $T=\sup D$ there is a sequence $t_n\to T$ with $t_n\in D$. Let $s_n=\tilde\sigma(t_n)$ be the corresponding sequence in $S$. As $S$ is compact $s_n$ has an adherence point $s$ in $S$. By continuity $f(s)=\sigma(T)$. The local diffeo property of $f$ now concludes that $T\in D$, whence $\bar D=D$ and $D$ is closed.

Solution 2:

Manifolds are compactly generated and a proper map between compactly generated spaces is closed. After establishing that $f$ is closed Sam's answer to this question completes the proof.