countable union of proper subspaces In an interview I was asked  to solve a question by using Baire Category Theorem (a complete metric space can not be written as union of nowhere dense subsets), the question was: 

"Is the vector space $\mathbb{R}^n$ can be written as countable union of its proper subspaces?"

My approach was: first I show that $\mathbb{R}^2$ cannot be written as countable union of straight lines passing through the origin and as lines in $\mathbb{R}^2$ are nowhere dense sets and $\mathbb{R}^2$ is a complete metric space, so I concluded from the Baire Category Theorem, and for higher dimension I claimed and showed that hyperplanes are also nowhere dense set but later I couldn't conclude the final result.
Could you please help me in this regard? Or any other way to solve this one? 
 A: Hints: 
a) It is well known that $\mathbb{R}^n$ is complete.
b) Every proper subspace of $\mathbb{R}^n$ is nowhere dense in $\mathbb{R}^n$. To prove this assume that there exist some proper subspace $V\subset\mathbb{R}^n$ such that is not nowhere dense in $\mathbb{R}^n$. Then there exist some ball $B(x,r)\subset\mathbb{R}^n$ such that $V$ is dense in $B(x,r)$. Without loss of generality we can assume that $x\in V$, (otherwise we can allways make a small shift). This means that for all $y\in B(x,r)$ and $\varepsilon>0$ we can find $v_y\in V$ such that $\|y-v_y\|\leq\varepsilon$. Take arbitrary $z\in\mathbb{R}^n$ and $\varepsilon>0$. Consider vector $y=x+r\|z\|^{-1}z$. Since $\|y-x\|<r$, then $y\in B(x,r)$. Hence we can find $v_y\in V$ such that $\|y-v_y\|\leq \varepsilon \|z\|^{-1}r$. Now consider $v_z=\|z\| r^{-1}(v_y-x)\in V$, then we get 
$$
\|z-v_z\|=\|z\|r^{-1}\|\|z\|^{-1}rz-(v_y-x)\|=\|z\|r^{-1}\|\|z\|^{-1}rz+x-v_y\|\leq
$$
$$
\|z\|r^{-1}\|y-v_y\|\leq\|z\|r^{-1}\varepsilon\|z\|^{-1}r=\varepsilon
$$
Thus for each $z\in \mathbb{R}^n$ and $\varepsilon>0$ we found $v_z\in V$ such that $\|z-v_z\|\leq\varepsilon$. This means that $V$ is dense in $\mathbb{R}^n$. Since $V$ is finite dimensional then it is closed. Since $V$ is closed and dense in $\mathbb{R}^n$, then $V=\mathbb{R}^n$. Contradiction, because $V$ is a proper subspace. Hence every proper subspace is nowhere dense in $\mathbb{R}^n$.
c) Now we apply Baire category theorem and get the desired result.
