If $f:M \to N$ is a local homeomorphism between manifolds and closed then the map $f$ is proper? A friend of mine asked me the following question: How do I prove that a local homeomorphism $f:M \to N$ between manifolds is closed iff it is proper?
It is well known and easy to prove that if $f$ is proper, then f is closed, but I'm not able to prove the converse and, being honest, I am not sure it is really true, but I also wasn't able to provide a counterexample.

Question: How do I prove that if  $f$ is closed, then it is proper?

 A: The result is true. Since the spaces $M$ and $N$ are manifolds, we may suppose they are metric spaces with metrics $d_M$ and $d_N$.
Take a compact $K \subset N$. We have the closed set $F =f^{-1}K$. Suppose $F$ is not compact. We have a sequence $x_n \in F$ that do not admit a convergent subsequence. Since $K$ is compact we may suppose $fx_n$ converges to a point $y\in K$. We can also suppose $x_n$ have no repeating term.
Notice $D = \{x_n \in X:n \in \mathbb N\}$ is discrete and closed, because $x_n$ do not admit convergent subsequences. By hypothesis, the map $f$ is closed and therefore $fD$ is closed. So, we must have $y \in fD$. 
Now let's analyse two possible cases:
First case: If $I = \{n \in \mathbb N: fx_n \neq y\}$ is an infinite set, then consider the closed set $D' = \{x_n \in X: n \in I\} \subset M$ and we have $fD' \subset N$ is closed. Since $y$ is a limit point of $fD'$, we have a contradiction.
Second case: If $I = \{n \in \mathbb N: fx_n = y\}$ is an infinite set, then for every $k \in I$ take a neighborhood $U_k$ of $x_k$ such that $fU_k$ is open and $f:U_k \to fU_k$ is homeomorphism. Reducing the open neighborhoods $U_k$ if necessary we may suppose $U_k \cap U_l = \emptyset$ if $k\neq l$. For each $k\in I$ take $z_k \in U_k \setminus \{x_k\}$ such that $d_M(z_k,x_k)<1/k$ and $d_N(fz_k,y)<1/k$. We have $D' = \{z_k \in X: k\in I\}$ is a closed discrete subset of $M$ and, therefore, $fD'$ is closed in $N$, which is a contradiction, because $y$ is a limit point of $fD'$ and $y \not \in fD'$.
Therefore, $F$ must be compact.
