Converge of distance between two sequences Let $(s_n)$ and $(t_n)$ be sequences in the metric space $(X,d).$ Suppose $s_n \rightarrow s \in X$ and $t_n \rightarrow t \in X.$ Prove that $d(s_n,t_n) \rightarrow d(s,t).$
My attempt: Assume $s_n \rightarrow s \in X$ and $t_n \rightarrow t \in X.$ Then $\forall \epsilon >0, \exists$ $N$ such that $n>N \implies d(s_n,s) < \frac{\epsilon}{2}.$ And  $\forall \epsilon >0, \exists$ $M$ such that $m>M \implies d(t_m,t) < \frac{\epsilon}{2}.$ Now let $K=max\{N,M\}$. Then $\forall \epsilon >0, \exists$ $K$ such that $n>K \implies d(s_n,t_n) \le d(s_n,s)+d(t_n,t)< \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$ 
Does this look correct? It's the only way I could think to try it. Thanks

You have the right idea, but didn't quite carry it out properly.
Hint: Work with triangle inequalities in two directions.
$$ d(s_n,t_n)\le d(s_n,s)+d(s,t)+d(t,t_n)$$
and
$$ d(s,t)\le d(s,s_n)+d(s_n,t_n)+d(t_n,t)$$
and see what happens when $n\to\infty$.
