A finite-dimensional vector space cannot be covered by finitely many proper subspaces? Let $V$ be a finite-dimensional vector space, $V_i$ is a proper subspace of $V$ for every $1\leq i\leq m$ for some integer $m$. In my linear algebra text, I've seen a result that $V$ can never be covered by $\{V_i\}$, but I don't know how to prove it correctly. I've written down my false proof below:
First we may prove the result when $V_i$ is a codimension-1 subspace. Since $codim(V_i)=1$, we can pick a vector $e_i\in V$ s.t. $V_i\oplus\mathcal{L}(e_i)=V$, where $\mathcal{L}(v)$ is the linear subspace span by $v$. Then we choose $e=e_1+\cdots+e_m$, I want to show that none of $V_i$ contains $e$ but I failed.
Could you tell me a simple and corrected proof to this result? Ideas of proof are also welcome~
Remark: As @Jim Conant mentioned that this is possible for finite field, I assume the base field of $V$ to be a number field.

This is a special case of a fact that an affine space over an infinite field is irreducible.
The proof can be found in most books on elementary algebraic geometry(see for example Fulton's algebraic curves).
