TITLE: Distance between a point and closed set in finite dimensional space
QUESTION [11 upvotes]: Let $X$ be a linear normed space.
I need to prove that $X$ is  finite dimensional normed space if and only if for every non empty closed set $C$ contained  in $X$ and for every $x$ in $X$ the distance $d(x,C)$ is achieved in specific $c$. I know how to prove the direction which assumes $X$ is finite dimensional (use Riesz lemma) but I dont know what to do in the other direction. 
thanks

REPLY [9 votes]: Suppose $X$ infinite dimensional. Then the unit sphere $S$ is not compact (Riesz theorem), and therefore there is a sequence $x = (x_n)_n$ on $S$ without accumulation points.
Denote by $x'$ the new sequence defined by $x_n' = (1 + \frac{1}{n})x_n$.
Since $x'$ and $x$ have the same accumulation points, $x'$ doesn't have any. So the set $C$ of the values of $x'$ is closed.
Now $d(0, C) = 1$, and there is no point in $C$ of norm $1$.