Let $X,Y$ denote arbitrary topological spaces. A map $f:X\to Y$ is called proper if for all compact $K\subset Y,$ the pre-image $f^{-1}(K)\subset X$ is also compact.
A net $\{x_\alpha\}_{\alpha\in A}\subset X$ goes to infinity if for all compact $K\subset X,$ there exists $\alpha_0\in A$ such that $x_\alpha\not\in K$ for all $\alpha\ge\alpha_0.$
I propose the following equivalent notion of properness: A map $f$ is proper if any only if for every net $\{x_\alpha\}_{\alpha\in A}\subset X$ that goes to infinity, the net $\{f(x_\alpha)\}_{\alpha\in A}\subset Y$ also goes to infinity.
Wikipedia asserts that this equivalency holds if we assume that $X,Y$ are metric spaces and we replace nets with sequences. I can prove one direction of the general characterization, namely the "only if" direction:
Proof by contrapositive. Let $\{x_\alpha\}_{\alpha\in A}\subset X$ denote a net that goes to infinity such that $\{f(x_\alpha)\}_{\alpha\in A}$ does not go to infinity. Let $K\subset Y$ denote a compact such that for all $\alpha_0\in A,$ there exists $\alpha\ge\alpha_0$ with $x_\alpha\in f^{-1}(K).$ Since $\{x_\alpha\}$ must eventually leave every compact, then $f^{-1}(K)$ cannot be compact.
As for the converse direction, I do not know. Here is one thing that might help: In any topological space $X$ with at least one compact subset that is not itself compact, I can construct a non-empty net that goes to infinity. Indeed, let the directed set $A$ be the collection of all compact subsets, ordered by inclusion. For all $K\in A,$ choose $x_K\in X$ so that $x_K\not\in K.$ Clearly $\{x_K\}_{K\in A}$ goes to infinity. I believe that a variation of this construction will do the trick for the converse direction.
Best Answer
The converse is not true in general. For instance, if $X$ is compact, then any map $f:X\to Y$ for any $Y$ vacuously satisfies the second condition, since no net in $X$ goes to infinity. But not every map $f:X\to Y$ is proper. For instance, if $Y$ is $X$ with the indiscrete topology and $f$ is the identity map, then every subset of $Y$ is compact so this would be saying every subset of $X$ is compact which is certainly not true in general.
It is true if you require $Y$ to be Hausdorff. In that case, suppose $f:X\to Y$ is not proper; let $K\subseteq Y$ be compact such that $f^{-1}(K)$ is not compact. We can then find some net $(x_\alpha)$ in $f^{-1}(K)$ with no accumulation point in $f^{-1}(K)$. Since $Y$ is Hausdorff, $K$ is closed, so $f^{-1}(K)$ is closed, and so $(x_\alpha)$ actually has no accumulation point in $X$. It follows that $(x_\alpha)$ must go to infinity. But $f(x_\alpha)\in K$ for all $\alpha$, so $(f(x_\alpha))$ does not go to infinity.