Suppose that a metric space has a countable dense cover . Show that any open cover of has a subcover which is at most countable.

So consider an arbitrary cover of . Let be the set of open balls where and . Let be the set of open balls such that is contained in one of the . From here can we say the these union of open balls is because of the denseness of ? From here how do we show that the subcover is either finite or countable?