Show there exists $v\in V$ such that $v\not\in V_{i}$ for any $1\leq i\leq k$ 
Show that if $V$ is a nonzero vector space over $\mathbb{R}$, and $V_{1}, V_{2},\ldots V_{k}$ are proper subspaces of $V$ then there exists $v\in V$ such that $v\not\in V_{i}$ for any $1\leq i\leq k$

I can prove the case for $k=2$ but I cannot produce an induction argument. 
 A: Let $V=V_1\cup V_2\cup \cdots \cup V_n$ where $V_1\not\subset V_2\cup V_3\cup \cdots \cup V_n$.
Let $v\in V_1\setminus( V_2\cup V_3\cup \cdots \cup V_n)$.
Also since $V_1$ is proper so $\exists w\in V\setminus V_1$.
Consider $P=\{w+rv:r\in \Bbb R\}$ which is definitely an infinite subspace of $V$.
Claim: $V_i$ contains at-most one member of $P$ for each $1\le i\le n$.
If $w+rv\in V_1$ for some $r\in \Bbb R\implies w\in V_1$ which is false.
If $|V_i|>1$ and for $r_1\neq r_2$$;w+r_1v,w+r_2v\in V_i;i\ge 2\implies (r_1-r_2)v\in V_i\implies v\in V_i$ which is false and hence $|V_i|<1$.
Hence $V_i$ contains at-most one member of $P$ for each $1\le i\le n\implies V$ contains finite number of members of $P$ which is false since $P$ is an infinite subspace of $V$.
Hence $V\neq V_1\cup V_2\cup \cdots \cup V_n$ ,hence there exists $v\in V$ such that $v\notin V_i$ for any $i$.
A: A proof using some topology in the finite dimensional case (in the general case and using only linear algebra, see learnmore's answer) : 
If each of the $V_i$ is proper then they're all closed sets with empty interior, which, according to the Baire Category theorem implies that their union also has empty interior : therefore it isn't $V$.
