You are quite correct that this is a purely topological statement. Proper maps are closed and here is a proof that works for closed maps between topological spaces, not necessarily manifolds nor even locally compact .
Proof: Since $U$ is open in $M$, $M \setminus U$ is closed, hence (by the assumption "closed map" ) so is $F\subset N$ defined by $F=f(M\setminus U) $. Just take $V=N\setminus F$. Done.
Edit: Since the issue of "properness" is a very interesting one, I hope you will excuse the following development, prompted by Theo's comment.
The principal property of proper maps is that they are closed and I think this should be built into the definition (obviously Algebraic Geometry makes its influence felt here). So a nice definition is that a continuous map $f:X\to Y$ between topological spaces is proper if it is closed and if the inverse images of points in $y \in Y$ are compact subsets $f^{-1}(y) \subset X$ [technically some say "quasi-compact" subsets to stress that no Hausdorff condition is assumed]. The link with the usual notion of "properness" is then the following theorem (in which proper has the meaning just given), whose assumptions are pleasantly weaker than one might expect. The point I advocate is that one should learn it once and for all, and that then you can often give short elegant proofs in the context of properness.
Theorem If $X$ is Hausdorff and $ Y $ locally compact, then $f$ is proper if and only if the inverse image of a compact subset $K\subset Y$ is a compact subset $f^{-1}(K) \subset Y$.
(The proof is in Chapter 1 of Bourbaki's book on General Topology, which adopts this point of view)