Let be $(X,d)$ a metric space, $G$ a group acting on $X$. The action is properly discontinuous if and only if for all $x \in X$ exists $k>0$ such that $ \{g \in G | d(x,gx) < k\} = \{e\}$.
$G$ is acting by isometries and properly discontinuous action is an action such that for all $x \in X$ exists a neighborhood U of $x$ such that, if $g \neq e$, $gU \cap U$ = empty. $ \\ $ $ \ \ $
I learning on my own and I found this observation and I can not prove it. I can not see why proper discontinuous action is equivalent to that in a metric space.
Suppose $G$ acts properly discontinuously and let $x\in X$. Let $U$ be a neigborhood of $x$ such that $gU\cap U=\emptyset$ for all $g\neq e$. Let $k>0$ be such that $U$ contains the ball of radius $k$ around $x$. Then if $g\neq e$, $gx\in gU$ so $gx\not\in U$. In particular, $d(x,gx)\geq k$. This shows $k$ satisfies your condition.
Conversely, suppose the action of $G$ satisfies your condition and let $x\in X$. Choose $k>0$ such that $\{g\in G:d(x,gx)<k\}=\{e\}$ and let $U$ be the ball of radius $k/2$ around $x$. Since $G$ acts by isometries, $gU$ is the ball of radius $k/2$ around $gx$ for each $g\in G$. If $g\neq e$, then if $y\in U$, $d(gx,y)\geq d(gx,x)-d(y,x)>k-k/2=k/2$. Thus $y\not\in gU$. Since $y\in U$ was arbitrary, $gU\cap U=\emptyset$. Thus the action is properly discontinuous.