$x \in \mathbb R^d$, $B$(closed)$ \subseteq \mathbb R^d$. Where $\mathbb R^d$ is an inner product space. Proof help
Let $B\subset\mathbb{R}^d$ be closed and $z\in\mathbb{R}^d$, show that there is $b_0\in B$ such that $\delta(z,B)=\|z-b_0\|$.