I would like to prove the following.
Let $M$ be a closed manifold. Then $\pi_1(M)$ acts properly discontinuously on its universal cover $\tilde{M}$.
Here I am using the definition of a properly discontinuous action to be that for all compact $K \subset \tilde{M}$, the set $\{ g \in \pi_1 (M) : gK \cap K \neq \emptyset \}$ is finite.
My incomplete attempt goes as follows. Assume for a contradiction that there exists a sequence of distinct elements $\{ g_i \}_{i=1}^\infty$ such that $ g_i K \cap K \neq \emptyset $. Then for all $g_i$ there exists some (distinct) $k_i$ such that $k_i \in g_i K \cap K$. Since in particular all such $k_i$ belongs in $K$, and it is an infinite set, there is a limit point $k$ inside $K$.
This is where I'm stuck. I would assume I take a small enough open neighborhood $U$ of $k$ and use properties of the covering map $p : \tilde{M} \to M$ to show that this neighborhood never intersects any $gU$ for all $g \in \pi_1(M)$ and this would yield a contradiction with the above argument in $K$, but I don't know how to flesh this out. I also know that the action is co-compact, if this is of any use.
Any help would be appreciated!