Since this old question does seem to be of some interest, here is a proof. First of all, here is a useful (but not well-known) definition.

**Definition.** Let $X$ be a Hausdorff 1st countable topological space, $G\times X\to X$ a topological action of a discrete topological group $G$. Two points $x, y\in X$ are said to be **dynamically related** with respect to this action if there is an infinite sequence (of distinct elements) $g_n\in G$ and a sequence $x_n\to x$ in $X$ such that $g_n(x_n)\to y$.

It is a pleasant exercise to show that an action which admits a pair of dynamically related points cannot be properly discontinuous. (Consider the compact set formed by $\{x, y\}$ and the image of the sequence $(x_n)$.) The converse is also true but is not relevant for the question.

Now, to the question itself. Suppose that $G\times X\to X$ be a properly discontinuous action. (A proper action is clearly not enough, as you can consider say, $U(1)$ acting on itself.) Suppose that $\epsilon>0$ as in the question does not exist. Then for every $n$ there is $g_n\in G$ (with $g_n$'s distinct for distinct $n$'s) and points $x_n, y_n\in C$ such that $d(y_n, g_n(x_n))<1/n$. After extraction, we can assume that $x_n\to x\in C, y_n\to y\in C$. Now, it is clear that $x, y$ are dynamically related. qed

Note that we did not need the action $G\times X\to X$ to be isometric.