A finite-dimensional vector space cannot be covered by finitely many proper subspaces?

[Edit] This answer is contained in another answer of mine. Sorry about that. Switching to CW[/Edit]

Pick a basis and a system of coordinates $x_1,\ldots,x_n$ for $V$. WLOG assume that $n \geq 2$. As you observed, without loss of generality we can assume that the subspaces are all of codimension one, i.e. spaces of solutions of a single homogeneous equation $$ a_1x_1+a_2x_2+\cdots +a_nx_n=0 $$ in the coordinates $x_i,i=1,\ldots,n$. Therefore a single subspace will intersect the infinite set $$ S=\{(1,t,t^2,\ldots,t^{n-1})\mid t\in k\} $$ at finitely many points, because the polynomial $a_1+a_2t+a_3t^2+\cdots+a_nt^{n-1}$ has at most $n-1$ zeros.

Therefore it is impossible to cover all of $S$, hence all of $V$, with finitely many subspaces.

Note that if $k$ is uncountable, then this argument shows that we need uncountably many subspaces.

Do you know the proof that the union of two subspaces of a vector space is a subspace if and only if one of the two subspaces is contained in the other? If the field is infinte one can come up with a similar proof for your statement.

Assume $V$ is covered by finitely many $V_i$, and assume that the cover is minimal. Then there is wlog a $v\in V_1$ which is not in any other $V_i$ and there is also wlog a $w\in V_2-V_1$. Then the vectors $av+w$ for $0\neq a\in k$ (where $k$ is the base field) are in pairwise different spaces $V_i$. Indeed if $av+w$ and $bv+w$ both are in $V_i$, then so is $(a-b)v$ which is a contradiction. Since $k$ is infinite this proves your statement.