Show that $\mathbb{R}^n$ cannot be written as a countable union of proper subspaces

Ok so I know I have to use Baire's Cathegory Theorem here. And I've done the following, lets suppose on the contrary that $\mathbb{R}^{n}=\bigcup_{n\in \mathbb{N}}S_n$ Such that $\forall n \in \mathbb{N}, S_n$ is a proper subspace. Now if I show that for every $n \in \mathbb{N}$, $S_{n}$ is closed and $int(S_{n})=\emptyset$ then Baire would imply $int(\mathbb{R}^n)=\emptyset$ which is a contradiction.

Now I was able to prove that for every $n \in \mathbb{N}$, $int(S_{n})=\emptyset$.

However, when I am trying to prove that $S_{n}$ is closed, I have the following inconvenient. $S_{n}$ is closed iff $\mathbb{R}^n-S_{n}$ is open. Therefore, if $\mathbb{R}^n-S_{n}$, is open given $x\in X$there exists $\epsilon>0$ such that $B(x,\epsilon)\subset \mathbb{R}^n-S_{n}$. Now given $a \in S_{n}$, we can consider $\frac{x-a}{\|x-a\|}\frac{\epsilon}{2}$ which satisfies $\frac{x-a}{\|x-a\|}\frac{\epsilon}{2} \in B(x,\epsilon)\subset \mathbb{R}^n-S_{n} $ Which implies $a \in \mathbb{R}^n-S_{n}$ which is a contradiction, and therefore $\mathbb{R}^n-S_{n}$ is not open.

So with this proof I can't use Baire! I don't know if this proof that shows that $S_{n}$ is not closed is ok and therefore my strategy to prove the exercise is wrong or if this latter proof is incorrect!

Suggestions to solve the problem in a different way are also more than welcome!