Infinite subset of a metric space has an infinite r-discrete subset or a non-trivial Cauchy sequence When working on functions on metric spaces I found I needed to confirm the following: For a metric $d$ on a space $X$ and and an infinite $Y\subset$$X$,there is an infinite $Z=\{a_n :n \in N\}\subset$ $Y$ where either $Z$ is $r,d$-discrete for some real $r>0$,or $(z_n)_{n\in N}$ is a $d$-Cauchy sequence. A set $Z\subset$$X$ is $r,d$-discrete iff  $d(p.q) > r$ whenever $p,q$ are distinct members of $Z$.( Note that the property of being an $r,d$-discrete set for some, or any, $r$ is not a topological property; it generally depends on the metric $d$.) I proved it using some infinitary combinatorics.I'd like to know if there is a more elementary proof. 
 A: I’ll take $d$ as given and omit it from the notation. Suppose that $Y$ contains no infinite $r$-discrete set for any $r>0$; then $Y$ is totally bounded. But a metric space is totally bounded if and only if every sequence in the space has a Cauchy subsequence, so $Y$ must contain a Cauchy sequence of distinct points.
Specifically, for $n\in\Bbb Z^+$ let $F_n$ be a finite subset of $Y$ such that $Y\subseteq\bigcup_{x\in F_n}B\left(x,\frac1n\right)$, and let $\sigma=\langle y_n:n\in\Bbb Z^+\rangle$ be a sequence in $Y$. There is an $x_1\in F_1$ such that $$N_1=\{n\in\Bbb Z^+:y_n\in B(x_1,1)\}$$ is infinite. Given an infinite $N_k\subseteq\Bbb Z^+$ for some $k\ge 1$, there is an $x_{k+1}\in F_{k+1}$ such that
$$N_{k+1}=\left\{n\in N_k:y_n\in B\left(x_{k+1},\frac1{k+1}\right)\right\}$$
is infinite. Now let $\langle y_{n_k}:k\in\Bbb Z^+\rangle$ be a subsequence of $\sigma$ such that $n_k\in N_k$ for each $k\in\Bbb Z^+$; clearly $\langle y_{n_k}:k\in\Bbb Z^+\rangle$ is Cauchy. The converse implication is straightforward.
