Take a discrete rank $n$ free abelian subgroup $\Gamma$ of the group of diagonal matrices in $SL(n+1, \mathbb R)$ with positive diagonal entries. The group $\Gamma$ will preserve each open coordinate orthant $C_i, i=1,...,2^{n+1}$ in $\mathbb R^{n+1}$. The projectivization $P(C_i)\subset \mathbb R P^n$ of each $C_i$ is open and properly convex: There will be $2^n$ of these domains in $\mathbb R P^n$. The group $\Gamma$ preserves each $P(C_i)$, acts on it properly discontinuously and cocompactly.
Edit: As for your update (which starts to sound like a fishing expedition). Suppose that $\Gamma$ is a hyperbolic group acting on a properly convex open subset $D\subset \mathbb R P^n$, equivalently, the closure of $D$ is strictly convex (the boundary does not contain any nondegenerate line segments). Then the boundary of $D$ is equivariantly homeomorphic to the Gromov-boundary of $\Gamma$ and, accordingly, the action of $\Gamma$ on the closure of $D$ satisfies the convergence property: Every unbounded sequence in $\Gamma$ contains a subsequence which converges uniformly on compact to constant in $\partial D$ away from a point in $\partial D$. In particular, each infinite order element of $\Gamma$ has a unique dominant eigenvector in $\mathbb R^{n+1}$ corresponding to its attractive fixed point in $\partial D$. Furthermore, attractive fixed points are dense in $\partial D$. Assume that $\Gamma$ acts
on two distinct properly convex domains $D_1, D_2$ in $\mathbb R P^n$ properly discontinuously and cocompactly. Then, by the above density observation, $\partial D_1=\partial D_2$. This is impossible for strictly convex domains.
