I have to prove the following :
Let G be a metric group acting continuously on X a metric space. Then the following are equivalent : (a) $G\curvearrowright X$ is proper.
(b) For all compact $K\subseteq X$ the set $C= \{g;g\cdot K\cap K\neq\emptyset\}\subseteq G$ is compact.
(c) For all converging sequences $(x_{n})_{n\in\mathbb{N}}\in X^{\mathbb{N}}$, for all diverging sequences $(g_{n})_{n\in\mathbb{N}}\in G^{\mathbb{N}}$, the sequence $(g_{n}\cdot x_{n})_{n\in\mathbb{N}}\in X^{\mathbb{N}}$ diverges.
I have managed to prove (a) $\Leftrightarrow$ (b), but I am stuck when it comes to prove (b) $\Rightarrow$ (c) or (c) $\Rightarrow$ (b). I don't see how the properness and the convergence of sequences are linked (I know that in a metric space, a definition of compactness is that any sequence has a converging subsequence, but I don't see how it helps).
I know that G and X are $\sigma$ locally compact because of the definition of a proper action. Since X is a metric space it is in particular hausdorff and hence any compact subspace of X is closed as well.
I tried to do it by contradiction. I assumed that the sequence $g_{n}.x_{n}$ was converging and then constructed a compact set $K \subset X$ so that $g_{n}\in C \forall n$. But I don't know how to do conclude from here that C is not compact from there.
Thank you very much beforehand for your help
For (c)->(b), suppose $C$ was not compact. Let $(g_n)$ be a sequence in $C$ without converging subsequence. By definition of $C$, choose $x_n\in K$ s.t. $g_nx_n\in K$. Going to subsequences from compactness of $K$, we can assume that $x_n$ and $g_nx_n$ converge, but $g_n$ diverges, which contradicts (C).