We have a proper map $f$, this means a continous map $f \colon X \to Y$, so that for every $K \subset Y$ compact, also $f^{-1}(K)$ is compact.
Question: Let $D_{1} \subset R^n$ and $D_{2} \subset R^m$ be bounded open sets (domains). Then the map $f \colon D_{1} \to D_{2}$ is proper if and only if for every sequence $\{x_{n}\} \subset D_{1}$ with the property $\{x_{n}\} \to \partial D_{1}$, we have $\{f(x_{n})\} \to \partial D_{2}$.
I think I got it in one direction. Lets say there exists a sequence $\{x_{n}\} \to \partial D_{1}$, such that $\{f(x_{n})\}$ does not converge to $\partial D_{2}$. Then $lim_{n \to \infty} f(x_{n}) = y \in D_{2}$. Lets take a compact $K$ around $y$ which contains the convergent sequence $\{f(x_{n})\}$. In its preimage $f^{-1}(K)$, there is the sequence $\{x_{n}\}$ which has no limits inside $f^{-1}(K)$, therefore $f^{-1}(K)$ is not compact. So we have found a compact $K$, whose preimage $f^{-1}(K)$ is not compact and so $f$ is not a proper map.
Any ideas for the other direction ?