If $\{s_n\}$ is a sequence and if $n_1, n_2, \cdots$ are positive integers such that $n_1 < n_2 < \cdots$, then the sequence $\{s_{n_{k}}\}$ is called a subsequence of $\{s_n\}$. Prove that a sequence in a metric space has limit $s$ if and only if every subsequence has limit $s$.
I know it has something to do with Cauchy sequences, I'm confused as to how I should go about this proof. Contradiction?
If part. Say, $d(s_{n},s)<\epsilon$ for $n\geq N$. For a subsequence $(s_{n_{k}})$, choose a large $K$ such that $n_{k}\geq N$ for all $k\geq N$, then $d(s_{n_{k}},s)<\epsilon$.