action of a subgroup on a metric space If $G$ acts properly and cocompactly by isometries on the metric space $X$
and if $H$ is a subgroup of $G$. Does $H$ act properly and cocompactly by isometries on a subspace of $X$?
 A: The fact that $H$ will still act by isometries is (trivially) true.
An action is proper if $f: G\times X\rightarrow X$ is proper, i.e. $f^{-1}(K)$ is compact for any compact $K\subseteq X$. Clearly if $H$ is a closed subgroup and $f$ is proper then $f_{|H\times X}^{-1}(K)$ will be compact because it is, as a set, $f^{-1}(K)\cap H\times X$, $X$ being a metric space, it is Hausdorff and hence any compact is closed so that $f_{|H\times X}^{-1}(K)$ is a closed set included in a compact set $f^{-1}(K)$ whence compact. 
As for the cocompactness, I cannot see any reason for this to hold in general. Take for instance $X=\mathbb{R}^2$ and $G=\mathbb{Z}^2$ acting by translation. The action is well known to be proper and cocompact (furthermore $X/G=\mathbb{T}$), however if $H$ is the trivial subgroup then $X/H=X$ and is not compact. If you consider the trivial group to be too particular you can also take $H:=\mathbb{Z}\times\{0\}$ in which case $X/H=S^1\times \mathbb{R}$ which is not compact as well.
One property that holds : if $H$ is a finite index subgroup of $G$ and $G$ acts properly and cocompactly on $X$ then $H$ will still act properly and cocompactly on $X$.
