I don't understand a step in the proof of the following proposition:
Let $\Lambda$ be a subgroup of a real vector space $V$ of finite dimension. Then $\Lambda$ is a full lattice if and only if $\Lambda$ is discrete and $V/\Lambda$ is compact.
Proof.(the part I don't understand) Assume $\Lambda$ is discrete and $V/\Lambda$ is compact, let $W$ be the subspace of $V$ spanned by $\Lambda$, then the real vector space $V/W$ can't have positive dimension since $V/\Lambda$ is compact.
Why is $V/W$ a trivial vector space?
The first thing going on here is a sequence of equivalencies: $\Lambda$ does not span $V \iff W$ is a proper subspace of $V \iff V/W$ has positive dimension $\iff V/W$ is a nontrivial vector space.
But if all of that happens then $V/W$ is noncompact. The inclusion $\Lambda \subset W$ induces a surjective continuous function $V / \Lambda \mapsto V / W$, and therefore $V / \Lambda$ is noncompact, a contradiction.