Let (an) be a convergent sequence in metric space X. Let "a" be
the limit of an. Then for sufficiently large N, if x,y>N,
d(ay,a)<e/2 and d(ax,a)<e/2 for any e greater than zero.
We add these inequalities and get d(ay,a) + d(ax,a)<e. But by
the triangle inequality d(ay,ax)<d(ay,a) + d(ax,a), so
d(ay,ax)<e for all x,y>N.
(some of those < should be less than or equal to, but it
doesn't really affect the proof.)