Suppose $S$ is any proper vector subspace of $\mathbb{R}^n$. Is $S$ a closed set in the usual topology on $\mathbb{R}^n$?
Geometrically, I think it is clear that $S$ must be closed in $\mathbb{R}^n$ for $1\leq n\leq 3$. If $x\in\mathbb{R}^n\setminus S$, then we can always find the shortest distance from $x$ to the point, line, or plane that is $S$ by taking the projection of $x$ onto $S$. Choosing $\epsilon$ to be less than this distance, $B_\epsilon(x)$ is an open ball around $x$ disjoint from $S$, so $S$ is closed.
I believe that the idea should carry over for higher dimensions, but I'm not sure how to make a more rigorous argument. How could this claim be proven without geometric handwaving? Thanks.