Just out of curiosity, is the following true?
If $U$ is a subspace of a normed vector space $X$, and $x\in X\setminus U$, then $\inf_{u\in U}\|x-u\|>0.$
If it is true, is it moreover true that there is $u_0\in U$ such that $\|x-u_0\|=\inf_{u\in U}\|x-u\|$?
If the norm is induced by inner product, this is clear because there is a notion of orthogonal projection.
Assuming $\inf_{u\in U}\|x-u\|=0$, then for all $n\in\mathbb{N}$ there exists $u_n$ such that $\|x-u_n\|<\frac1n$. Therefore $\lim u_n=x$. So we get a contradiction if $U$ is closed.
I can't see any obvious contradiction in the general case. My imagination is limited to $\mathbb{R}^n$ $(n=1,2,3)$, so I can't come up with a counterexample (in which $U$ is not closed and hence $X$ is not finite dimensional).