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.

I know that sequentially compact and compact are equivalent for metric spaces. So in order to show it is not compact I plan on showing it is not sequentially compact (meaning that not every sequence in $X$ has convergent subsequence). So I need to show that $(x_n)$ does not have a convergent subsequence, but how would I go about doing that? Or should I be attempting a completely different approach for my proof?

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    $\begingroup$ What is a famous necessary condition for a sequence to converge? $\endgroup$ – Daniel Fischer Dec 1 '13 at 21:37

You approach is correct. Show that the given sequence does not have a convergent subsequence, for which you can follow the next steps:

1) Assume that the sequence does have a convergent subsequence $y_n=x_{k_n}$.

2) The sequences $y_n$ then satisfies: $d(y_n,y_m)\ge r>0$.

3) and also, for every $\epsilon >0$ there exists $N_0$ such that .... (you have a choice here between two valid ways to fill in for the dots).

4) Now, make a particularly suitable choice for an $\epsilon$ above (use $r$) to obtain a contradiction.

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