By a standard theorem, there is a basis $B=\{b_\alpha\}$. Enumerate a countable subset of this and call it $\{b_i\}$. Let $F(D)$ denote all linear combinations of $D$ over the field $F$, for an arbitrary set of vectors $D$. Now consider the subspaces given by
$W_1=F(B-b_1)\\ W_2=F(B-b_2)\\ W_3=F(B-b_3)\\...$
There are a countable number of these, and it remains to show their union is all of $V$. But any $v\in V$ can be written as a finite linear combination of elements in $B$. Let the set of elements in $B$ used in this combination be $B_0$. Because $B_0$ is finite, there exists $b_i\in B$ with $b_i\notin B_0$. (If $B_0$ contained all of the $b_i$, it would be infinite.) Then $v\in W_i$, and we are done.