Group acts on a topological space Group $G$ acts cocompactly by isometries on a metric space $X$. Additionaly, for every point $x \in X$ there exists a neighbourhood $x \in U$  such that $g \cdot U \cap U = \emptyset$ for all but finitely many $g \in G$ (i). How can I show that if $K \subset X $ is compact then $g \cdot K \cap K = \emptyset$ for all but finitely many $g \in G$(ii)? 
All I can do is assume that (ii) is not true for some compact $K$ and then show that there exist two neighbourhoods from property (i) - $U_1,U_2$ such that $g\cdot U_1 \cap U_2 $ is not empty for infinitely many $g \in G$. Thank you for any answers.
 A: In fact, we don’t need the cocompacteness of the action.
First we claim that for every points $x$ and $y$ of the metric space $(X,d)$ there exist neighborhoods $U$ of $x$ and $V$ of $y$  such that $g \cdot U \cap V = \emptyset$ for all but finitely many $g \in G$ (iii). Indeed, (i) implies that the stabilizer $G_x=\{g\in G: G_x=x\}$ is finite and the orbit $G/G_x(x)$ is uniformly discrete, that is there exists a number $\varepsilon>0$ such that for each elements $g,h\in G$ such that $g(x)\ne h(x)$ holds $d(g(x),h(x))\ge\varepsilon$.  Indeed, assume the converse, there exists a sequence ${(g_n,h_n)}$ of pair of elements of the group $G$ such that $g_n(x)\ne h_n(x)$ and  $d(g_n(x),h_n(x))<1/n$. Since $G$ acts by isometries, we obtain that $ d(x, g_n^{-1}h_n(x))=d(g_n(x),h_n(x))<1/n$ for each $n$, which contradicts to (i). Now it suffices to put $U=\{x’\in X:d(x,x’)<\varepsilon/4\}$ and $V=\{y’\in X:d(y,y’)<\varepsilon/4\}$. Indeed, assume that there exist
elements  $g_1, g_2$ of the group $G$ and $x_1, x_2$   of the neighborhood $U$ such that both $g_1(x_1)$ and $g_2(x_2)$ belong to $V$. Since both $g_1$ and $g_2$ are isometries, we have that 
$$d(g_1(x),g_2(x))\le d(g_1(x),g_1(x_1))+ d(g_1(x_1),y)+ d(y,g_2(x_2))+ d(g_2(x_2),g_2(x))<$$ $$\varepsilon/4+\varepsilon/4+\varepsilon/4+\varepsilon/4=\varepsilon.$$
That is $g_1(x)=g_2(x)$, therefore $g_1^{-1}g_2\in G_x$. It remains to recall that the group $G_x$ is finite.   
I wonder, whether Claim (iii) holds for an arbitrary (Hausdorff) topological space $X$ and a group $G$ acting on $X$ by homeomorphisms and satisfying (i) and this question seems for me much more hard and interesting than the initial.
Now suppose that (ii) doesn’t hold. There exists a countably infinite subset $C$ of $G$ such that $gK\cap K\ne\varnothing$ for each element $g\in C$. For each such element fix points $(x_g,y_g)\in K\times K$ such that $gx_g=y_g$. Since the set $K\times K$ is (countably) compact, the set $\{(x_g,y_g):g\in C\}$ has a cluster point $(x,y)$. Then for each neighborhoods $U$ of $x$ and $V$ of $y$ we have $g \cdot U \cap V\ne\emptyset$ for infinitely many $g \in C$, a contradiction with (iii).
