A proof question about continuity and norm 
Let $E⊂\mathbb{R}^{n}$ be a closed, non-empty set and $N : \mathbb{R}^{n}\to\mathbb{R}$ be a norm. Prove that the function $f(x)=\inf\left \{ N(x-a)\mid a\in E \right \}$, $f : \mathbb{R}^{n}→\mathbb{R}$ is continuous and $f^{-1}(0) = E$.

There are some hints:
$f^{-1}(0) = E$ will be implied by $E$ closed, and $f : \mathbb{R}^{n}\to\mathbb{R}$ is continuous will be implied by triangle inequality.
I still can't get the proof by the hint. So...thank you for your help!

The other answers so far are good, but here is an alternative hint for the first part.
Because $E$ is closed, its complement $E^c$ is open. A set in $\mathbb{R}^n$ is open if and only if the set contains an open ball around any point in the set. Thus, for any $x\in E^c$, there is some $r>0$ such that the open ball $B(x,r)\subset E^c$. What does that tell you about the minimum distance from $x$ to $E$?
