Let $(M, d)$ be a complete metric space and $A\subseteq M$. I want to show that $A$ is relatively compact (i.e., $\overline A$ is compact) iff there exists no infinite subset $S\subseteq A$ such that $\inf_{x,y\in S, x\neq y}d(x,y)>0$.
This is apparently "well-known", but I am not sure about my solution:
"$\Rightarrow$": Assume that there is an infinite $S\subseteq A$ such that $\inf_{x,y\in S, x\neq y}d(x,y)>0$. Pick a sequence $(s_n)_{n\in\mathbb N}$ in $S$. Since $\inf_{n,m\in \mathbb N, s_n\neq s_m}d(s_n,s_m)>0$ by assumption, we found a sequence in $A$ with no convergent subsequence.* Hence $A$ cannot be relative compact. (correct?)
"$\Leftarrow$": Suppose that there exists no infinite subset $S\subseteq A$ such that $\inf_{x,y\in S, x\neq y}d(x,y)>0$. That is, for each infinite $S\subseteq A$ it holds that $\inf_{x,y\in S,x\neq y}d(x,y) = 0$. Now take any sequence $(a_n)_{n\in\mathbb N}$ in $A$, and put $S := \{a_n : n\in\mathbb N\}$. By assumption, $\inf_{n,m\in\mathbb N,a_n\neq a_m}d(a_n,a_m) = 0$, so there exists a convergent subsequence of $(a_n)_{n\in\mathbb N}$.** That is, $A$ is relatively compact.
`* Assume there exists a convergent subsequence $(s_{\phi(n)})_{n\in\mathbb N}$. Then for each $\epsilon>0$ there is an $N\in\mathbb N$ such that $d(s_{\phi(n)},s_{\phi(m)}) < \epsilon = 0 + \epsilon$. But this means $\inf_{n,m\in\mathbb N,s_{\phi(n)}\neq s_{\phi(m)}} d(s_{\phi(n)}, s_{\phi(m)}) = 0$. A contradiction.
`** Assume that $\inf_{n,m\in\mathbb N,a_n\neq a_m}d(a_n,a_m) = 0$ and $(a_n)_{n\in\mathbb N}$ has no convergent subsequence. $\leadsto$ stuck: I think I have to invoke completeness here?
