I've been reading a proof concerning S-arithmetic subgroups of algebraic groups and I'm having trouble determining why the following step should be true. First, the setup:

Let $G$ be a connected absolutely simple adjoint algebraic group defined over a nonarchimedean local field $F$. Let $K \subset F$ be a global field and let $v_0$ be the pullback of the normalized valuation on $F$ using the embedding $K \hookrightarrow F$. Let $S$ be a set of discrete valuations on $K$ such that $S$ contains all archimedean places and does not contain any place where $G$ is anisotropic. Let $\mathcal{O}_K(S)$ be the ring of $S$-integers in $K$. Let $\Gamma = G(\mathcal{O}_K(S))$.

Let $\displaystyle G_S = \prod_{v \in S} G(K_v)$.

We are given that $\Gamma$ is discrete in $G(K_{v_0})$. We know from Margulis' "Discrete Subgroups of Semisimple Lie Groups," Theorem 3.2.5, that $\Gamma$ is a lattice in $G_S$.

The paper I'm reading concludes that $G(K_v)$ must be compact for all $v \in S \setminus \{ v_0 \}$. The thing that I'm having difficulty understanding is how this follow from the discreteness of $\Gamma$ and the fact that $\Gamma$ is a lattice in $G_S$. Can someone point me in the right direction?