# Show that the metric space is not compact

I have a proof that I would like some hints in solving:

Let $X$ be a metric space. Show, if there is an $r > 0$ and a sequence $(x_n)$ from $X$ such that $d(x_n,x_m) \geqslant r$ for $n≠m$, then $X$ is not compact.

Hint. If $X$ is compact then $(x_n)$ has convergent subsequence. In particular it is Cauchy.
• @JimDarson That is, there is increasing sequence $(n_k)$ of natural numbers satisfy that $\lim_{n\to\infty} a_{n_k}\to L$ for some $L$. The proof of `every convergnet sequence is Cauchy' needs the triangle inequality. – Hanul Jeon Oct 26 '13 at 16:01
HINT: Let $A$ be the closure of the $\{x_n\mid n\in\Bbb N\}$. Show that $\{X\setminus A\}\cup\{B(x_n,r/2)\mid n\in\Bbb N\}$ is an open cover of $X$ which does not have a finite subcover.