I think you will miss the notorious Thompson's group $F$: it acts properly isometrically on a $CAT(0)$ cube complex (see Dan Farley, http://www.users.muohio.edu/farleyds/Far1.pdf); but it cannot act properly isometrically on a finite-dimensional complex, as this would make it of finite cohomological dimension.
EDIT: Indeed as Mark pointed out, the action of $F$ on this $CAT(0)$-cube complex is NOT co-compact, hence my answer should be discarded. Don't trust MO too much.

