Properly discontinuous action in metric spaces Let $X$ be a metric space and $G$ a group of homeomorphisms. I have to prove the following equivalences:
1) $G$ acts properly discontinuous on $X$ (i.e. any $G$-orbit is discrete and any stabilizer is finite);
2) For any compact $K$ in $X$, $T(K) \cap K \neq \emptyset$ for finitely many $T \in G$;
3) Any point $x \in X$ has a neighborhood $V$ such that $T(V) \cap V \neq \emptyset$ implies $T(x)=x$.
I showed $2) \implies 1)$ as follows: if we assume by contradiction that some orbit is not discrete, the it admits a limit point $z_0$, thus any neighborhood of $z_0$ meets infinitely many of its images under $G$. So, if I take any compact subset $K$ containing $z_0$ and I pass to some open neighborhood contained in $K$, 1) must follow in order to avoid contradictions with the hypotheses in 2).
Moreover, I showed $1) \implies 3)$ in this way: just consider a point $x \in X$. Since its orbit is discrete, we will always find a ball $B_\epsilon(x)$ containing only $x$ (and no other points of the orbit). Just take $V \subseteq B_{\frac{\epsilon}{2}}$. Then the condition $T(V) \cap V \neq \emptyset$ implies necessarily that $T(x)=x$.
I'm trying to prove by contradiction $3) \implies 2)$, but I have no clue. Can you help me, please?
 A: It's no wonder you're having trouble proving that (3) implies (2), because it's not true.  Here's a counterexample. Let $X=\mathbb R^2\smallsetminus\{(0,0)\}$, and let $G$ be the group of homeomorphisms of $X$ defined by $T_n(x,y) = (2^n x, 2^{-n} y)$ for $n\in\mathbb Z$. This satisfies properties (1) and (3), but not (2). 
To see that (2) fails, consider the compact subset $K= \{(x,y): \max(|x|,|y|)=1\} \subseteq X$. (It's a square of side 2 centered at the origin.) For every positive $n$, the element $(1,2^{-n})$ lies in $K$, and it's also equal to $T_n(2^{-n},1)\in T_n(K)$. Thus $T_n(K)\cap K \ne \emptyset$ for infinitely many $n$.
For a discussion of the different definitions of "properly discontinuous," see my answer to this question.
ADDED:
It's also not true that (3) implies (1). A counterexample is given by letting $X = \mathbb R$, and $G$ be the group of homeomorphisms given by $T_n(x) = 2^n x$ for $n\in \mathbb Z$. This satisfies (3), but it doesn't satisfy (1) because the stabilizer of the origin is not finite.  
To address your attempted proofs:
There's a flaw in your proof that (2) $\implies$ (1): there might not be any compact set that contains a neighborhood of $z_0$. (This would be true if the metric space were locally compact, but that's not part of the hypothesis.) You have to be more careful about choosing the compact set $K$. Other than that, I think this basic outline can be turned into a proof.
There's also a gap in your proof that (1) $\implies$ (3): I don't see why $T(V)\cap V \ne \emptyset$ implies that $T(x)=x$.  If we assume that $T(V)\cap V \ne\emptyset$, this just means there is some $y\in B_{\epsilon/3}(x)$ such that $T^{-1}(y)$ is also in $B_{\epsilon/3}(x)$. You'd like to conclude from this that $T(x) \in B_\epsilon(x)$, but I don't see how to conclude this unless all the homeomorphisms in $G$ preserve the metric, which is not part of the hypothesis. To tell the truth, I'm not sure at the moment whether (1) $\implies$ (3) or not.
Are you sure you copied the question correctly?  If you add the hypothesis that $G$ consists entirely of isometries, then I think it's true that all three properties are equivalent.
