Here are two key facts:
1) We are working inside $\mathbb{R}^n$, in which a subset is compact iff it is closed and bounded (Heine-Borel).
2) A discrete subgroup $G$ of any Hausdorff topological group is closed (a proof of this was given recently on this site). More generally, any locally compact subgroup of a Hausdorff group is closed.
So let $G$ be a discrete subgroup of $\mathbb{R}^n$. We want to prove that for every ball $B$ of finite radius, $B \cap G$ is finite. We easily reduce to the following special case: for all $r \geq 0$, the intersection of $G$ with the closed ball $B_r$ of radius $r$ centered at $0$ is finite.
Now let us apply the above facts: since $B_r$ is closed and bounded in $\mathbb{R}^n$, it is compact. Since $G$ is closed in $\mathbb{R}^n$, $G \cap B_r$ is closed in $B_r$, so $G \cap B_r$ is compact. Moreover, $G \cap B_r$ is a subspace of the discrete space $G$, so it is discrete. But a topological space which is both compact and discrete is necessarily finite.