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!
 A: $f(x)=0$ implies that there is a sequence $a_k \in E$ with $N(x-a_k)\to 0$. So $a_k\to x$ and hence $x$ is in the closure of $E$.
To prove that $f$ is continuous try to estimate $|f(x)-f(y)|$. Notice that $N(y-a) \le N(x-a)+N(x-y)$ so that $f(y) \le f(x) + N(x-y)$. Swap $x$ and $y$ so that you get $|f(x)-f(y)|\le N(x-y)$. So, if $x\to y$ you know that $N(x-y)\to 0$ and hence $f(x)\to f(y)$.
A: I denote the norm by $d$,by Triangle inequality for $x\in \mathbb{R}^n$ we have for all $z\in E$ and for all $y\in\mathbb{R}^n$ $$f(x)=\inf_{z'\in E} d(x,z')\le d(x,z)+\le d(x,y)+d(y,z)$$ since $z$ was arbitrary we must have that $$f(x)\le d(x,y)+ \inf_{z\in E} d(y,z)=d(x,y)+f(y)$$ so $$f(x)-f(y)\le d(x,y)$$ reversing the role of $x$ and $y$ we obtain $$f(y)-f(x)\le d(x,y)$$
 hence we have $$|f(x)-f(y)\le d(x,y)$$ This is precisely uniform continuity as $\forall \epsilon>0$ and $x,y\in\mathbb{R}^n$ let $\epsilon=\delta$ Then $d(x,y)<\delta=\epsilon\Rightarrow |f(x)-f(y)|<\epsilon=\delta$
A: 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 $E$?
