First, let me fix some definitions.
The action of a group $G$ on a topological space $X$ is proper if for every compact subspace $K \subseteq X$ the set $\{g \in G \ | \ g K \cap K \neq\varnothing \} $ is finite.
If $X$ has some additional combinatorial structure (say a triangulation or a cubulation...), then $G$ is said to act combinatorially if it sends vertices to vertices and the action of higher-dimensional cells is determined by the action on vertices. Notice that the combinatorial structure is not assumed to be locally finite.
Now let $G$ be a discrete group acting combinatorially on $X$. If the action is proper of course vertex stabilizers are finite. I would like to see a proof of the reverse implication.
So I take a $K$ compact. Since vertex stabilizers are finite and a compact contains only finitely many vertices, only a finite number of $g\in G$ can fix a vertex in $K$; so I can reduce to consider elements that do not fix vertices in $K$.
If $g$ fixes some point $p$ which is not a vertex, then it fixes (setwise) the whole least-dimensional cell $C$ in which the point sits (because the action is combinatorial). As a result, such a $g$ induces a permutation of the vertices of $C$ (which form of course a finite set). Anyway I think different elements can induce the same vertex permutation, so finiteness is not guaranteed here, so I don't know how to conclude.
I should also consider elements without fixed points in $K$ but such elements "move everything enough": either a non-fixed point sits in a fixed cell (and we're back to the previous case), or it sits in a non-fixed cell, and in this case it is moved to another cell of the same dimension which is at least 1 apart (with respect to the natural combinatorial metric on $X$ - also notice $G$ acts isometrically with respect to this metric), again because the action is combinatorial and we're assuming there is no fixed point in this case.
I'd really appreciate any suggestions or comments.
There is some ambiguity about what is properness if you don't make more explicit the choice of topology and definition of properness (metric? topological?).
If you have a connected graph and deal with metric properness (which is the most relevant option in geometric group theory), the result is certainly not true: e.g. pick any infinite group $\Gamma$, endow it with the structure of a complete oriented graph. Then the stabilizers are trivial, but since this is a bounded space, the action is certainly not metrically proper.
If you have a connected graph and deal with topological properness with respect to the inductive limit topology, then any compact subset is contained in a finite union of edges, and properness easily follows. Also if you deal with topological properness, any compact subset is contained in the $(1/3)$-neighborhood of a finite union of edges and properness follows once more.