MHB A proof question about continuity

[i_a_n]

Let $E⊂\mathbb{R}^{n}$ be a closed, non-empty set and $\mathbb{R}^{n}→\mathbb{R}$ be a norm. Prove that
the function
$f(x) = inf$ {$N(x-a) s.t. a∈E$}, $f :\mathbb{R}^{n}→\mathbb{R}$ is continuous and $f^{-1}(0)=E$.(There are some hint:
$f^{-1}(0)=E$ will be implied by $E$ closed. $f :\mathbb{R}^{n}→\mathbb{R}$ is continuous implied by triangle inequality.I still can't get the proof by the hint. So...thank you for your help!)

 

[Jhoneine]

Proof:First, we prove that $f^{-1}(0)=E$. Let $x\in E$. Then, for all $a \in E$, $N(x-a) = 0$. Thus, we have $f(x) = inf$ {$N(x-a) s.t. a∈E$} $= 0$. Hence, $x\in f^{-1}(0)$. Conversely, let $x \in f^{-1}(0)$. Then $f(x) = 0$. This implies that for all $a \in E$, $N(x-a) = 0$. Since $E$ is closed, we have $x \in E$. Therefore, $f^{-1}(0)=E$. Next, we prove that $f :\mathbb{R}^{n}→\mathbb{R}$ is continuous. Let $x_{0}$ be any point in $\mathbb{R}^{n}$. Let $y_{0} = f(x_{0})$. Let $\epsilon > 0$ be any real number. We need to show that there exists a $\delta > 0$ such that for all $x \in \mathbb{R}^{n}$, if $N(x-x_{0}) < \delta$ then $|f(x) - y_{0}| < \epsilon$. Since $y_{0} = f(x_{0}) = inf$ {$N(x_{0}-a) s.t. a\in E$}, there exists an $a_{0}\in E$ such that $N(x_{0}-a_{0}) < y_{0} + \frac{\epsilon}{2}$. By the triangle inequality, we have $N(x-a_{0}) \leq N(x-x_{0}) + N(x_{0}-a_{0})$. Let $\delta = \frac{\epsilon}{2} + N(x_{0}-a_{0})$. Then for all $x \in \mathbb{R}^{n}$, if $