Equivalence to properly discontinuous action Let $X$ be a metric space and let $G$ be a group of homeomorphisms $X \to X$ acting on $X$. We say $G$'s action is properly discontinuous in case for every $x \in X$ and compact $K \subseteq X$, there are at most finitely many $g \in G$ such that $g(x) \in K$. Equivalently (and this is not hard to show), $G \cdot x$ is discrete and $G_x$ finite for any $x$.

Why is it the case that $G$ acts properly discontinuously if and only if for any compact $K$, $g(K) \cap K \ne \emptyset$ for only finitely many $g$? One direction is relatively easy, but I just cannot seem to prove the "only if". The best I've been able to do is for finite $K$ (which is kind of the next best thing when you're stumped on proving something for compact sets, I guess).
I taken a look in Svetlana Katok's "Fuchsian groups", and I'm really confused.

For instance, it is proved in that book that $G$ acts properly discontinuously on $X$ iff every point $x \in X$ has a neighborhood $U_x$ such that $U_x \cap gU_x$ is not empty for only finitely many $g \in G$. (I don't understand the "only if" part, and I don't think that this holds in general).

Although, let be $G$ the group of all the (continuous) transformations $f_n$, $n \in \mathbb{Z}$,  of the plane without the origine $(0,0)$, with $f_n:(x,y) \mapsto (2^nx,2^{-n}y)$.

One can easily see that for each point $(x,y)$, a disc with this point as a center and a sufficiently small radius   intersects the orbit of $(x,y)$ only in one point, thus the orbit of each point is a discret subset, and clearly G acts freely on our set, thus the stabilizer of each point is finite. 

On the other hand, consider the segment $K = \{(t, 1-t)| 0 \leq t \leq 1 \}$, clearly $K$ is compact, $K$ contains for each $n&gt;0$ the element $P_n =(\frac{2^n-1}{2^{2n}-1}, \frac{2^n (2^n-1)}{2^{2n}-1}) $, and $f_n(P_n) \in K\cap f_nK$.  Therefore $K \cap f_nK$ is not empty for infinitely many elements $f_n$ of $G$.
Suppose there is a compact $K$ such that $g(K) \cap K \neq 0$  for infinitely many $g$. Denote by $L_k := g_k(K) \cap K$ where $g_k, k\in \mathbb N$ are among those elements of $G$ for which the inequality holds. Now pick any sequence $y_k \in L_k$. Because $K$ is compact, we have a convergent subsequence $y_{k_n} \to y$. But because every $y_{k_n} \in g_{k_n}(K)$ there must also be a sequence $x_{k_n}$ in $K$ such that $g_{k_n} \cdot x_{k_n} = y_{k_n}$ and so again by compactness there is a convergent subsequence $x_{k_{n_m}} \to x$ of $x_{k_n}$. But then $g_{k_{n_m}}\cdot x \to y$ is not discrete.