Let $B\subset\mathbb{R}^d$ be closed and $z\in\mathbb{R}^d$, show that there is $b_0\in B$ such that $\delta(z,B)=\|z-b_0\|$. If there is a closed subset $B$ of $\mathbb{R}^d$ and $z\in \mathbb{R}^d$ then show there is a point $b_0\in B$ so that we have $\delta(z,B) = \|z-b_0\|.$
I'm not sure how to begin this exercise and what theorems are available to help me solve it
 A: Theorem of Heine-Borel: In finite dimensional spaces: compact = closed+bounded
Thus find an appropriate finite portion of the set B.
A: Let $(x_n)_n\subset B$ a sequence such that $\lim_{n\to\infty}\|x_n-z\| = \delta(z,B)$ (exists by definition of the infimum in $\delta(z,B)$). Let $\epsilon >0$, then there exists $N>0$ such that $$\Big|\|x_n-z\|-\delta(z,B)\Big|<\epsilon \qquad \forall n>N.$$
Then we have $$\|x_n\|\leq\|x_n-z\|+\|z\| \leq\|z\|+\delta(z,B)+\epsilon \qquad \forall n >N.$$ Thus the sequence $(x_n)_n$ is bounded by $C = \max\{\|x_1\|,\ldots,\|x_N\|,\|z\|+\delta(z,B)+\epsilon\}$. So, there is a subsequence $(x_{n_k})_k\subset (x_n)_n$ converging to some $b_0\in\Bbb R^d$. Since $B$ is closed and $(x_{n_k})_k\subset B$, we must have $b_0\in B$. Finally, since $\big(\|x_n-u\|\big)_n$ is converging to $\delta(z,B)$ it is also the case of the subsequence $\big(\|x_{n_k}-u\|\big)_{k}$. It follows by continuity of the norm that
$$\|b_0-z\| = \lim_{k\to \infty}\|x_{n_k}-u\| = \delta(z,B).$$
