Let $(M,d)$ be an unbounded metric space and $\delta>0$. Prove that $M$ has a $\delta$-skeleton, i.e., a subset $S$ of $M$ that satisfies: (1) $d(x,y)\geq \delta, \forall x,y\in S$; (2) $\forall x\in M, \exists u\in S$ s.t. $d(x,u) \leq \delta$.
My teacher said Zorn's lemma will help but I still don't know how to prove it.
Thank you.
Let $\mathscr S \subset P(M)$ be given by
$$\mathscr S = \{ A \subset M : d(x, y) \ge \delta \text{ for all } x, y \in A, x\neq y\}.$$
Note that $\mathscr S$ is partially ordered by the set inclusion. Let $A_n \in \mathscr S$ and $A_{n} \subset A_{n+1}$ for all $n$. Then
$$ A = \bigcup_{n=1}^\infty A_n$$
is still in $\mathscr S$: if $x, y\in A$ and $x\neq y$, then $x\in A_{n_1}$ and $y\in A_{n_2}$ for some $n_1, n_2$. Then $x, y\in A_m$ for $m = \max\{ n_1, n_2\}$ and thus $d(x, y) \ge \delta$.
Thus the Zorm lemma implies that $\mathcal S$ has a maximal element $S$. If $z\in M$ and $d(x, y) > \delta$ for all $y\in S$, then $S\cup \{z\}$ is also a $\delta$-skeleton, which is a contradiction to the choice of $S$.