I was reading the article "The Geometries of 3-manifolds" by Peter Scott and in the end of page 406 he states the following:
If $G$ acts properly discontinuously on a space $X$, then $G$ is a
discrete subset of the space of all continuous functions $X \to X$
with the compact-open topology. The converse is false, in general, but
is true if $X$ is a complete Riemannian manifold and $G$ is a group of
isometries of $X$.
How do I prove the last statement?
If $X$ is a complete Riemannian manifold and
$G$ is a group of isometries of $X$ acting properly discontinuously on $X$, then $G$ is discrete.
Observation: We say a group $G$ of homeomorphisms of $X$ acts properly discontinuously on $X$ if for every compact $K \subset X$ the set
$$\{g \in G: gK \cap K \neq \emptyset\}$$
is finite.
Best Answer
Suppose that $X$ is a (complete) metric space which satisfies the Heine-Borel property (every closed and bounded subset is compact). For instance, you can take $X$ to be a complete connected (finite dimensional) Riemannian manifold equipped with Riemannian distance function.
Then Arzela-Ascoli theorem implies that for every sequence of isometries $f_i: X\to X$ such that there exists $p\in X$ and $R$ for which $d(p, f_i(p))\le R$ for all $i$, there exists a subsequence $(f_{i_j})$ which converges to an isometry uniformly on compacts in $X$.
Given this, let us prove
Lemma. Suppose that $\Gamma$ is a discrete subgroup of $Isom(X)$ (the isometry group of $X$) equipped with the topology of uniform convergence on compacts. Then $\Gamma$ acts properly discontinuously on $X$.
Proof. Suppose not. Then there exists a compact $K\subset X$ and an infinite sequence of distinct elements $\gamma_i\in\Gamma$ such that $\gamma_i K\cap K\ne \emptyset$. Taking $p\in K$ and $R=2diam(K)$, we conclude that for each $\gamma_i$, $d(\gamma_i(p), p)\le R$. Therefore, by the above observation, $(\gamma_i)$ contains a convergent subsequence $(\gamma_{i_j})$. Taking the sequence of products $$ \alpha_j:= \gamma_{i_j}^{-1} \gamma_{i_{j+1}}, $$ we conclude that $\alpha_j\to id$ uniformly on compacts. (I am using here the property that $Isom(X)$ with topology of uniform convergence on compacts is a topological group.) Hence, $\Gamma$ is not a discrete subgroup of $Isom(X)$. A contradiction. qed