Let X be a metric space.
Every open cover in
has a countable subcover => a countable dense
set
{B(1/n, x) | all x in X} is clearly an open cover, whatever n>0 one
picks. for a fixed n, say n=2, there is a subcover
by the assumption. To get a dense set, we just
include xmn for all m and n.
The converse also holds.