Suppose that a metric space $\displaystyle X $ has a countable dense cover $\displaystyle Q $. Show that any open cover of $\displaystyle X $ has a subcover which is at most countable.

So consider an arbitrary cover of $\displaystyle X $ $\displaystyle , \{U_{\alpha}: \alpha \in A \} $. Let $\displaystyle \mathcal{B} $ be the set of open balls $\displaystyle B_{r}(q) $ where $\displaystyle r \in \mathbb{Q} $ and $\displaystyle q \in Q $. Let $\displaystyle E \subset \mathcal{B} $ be the set of open balls such that $\displaystyle U_{\alpha} $ is contained in one of the $\displaystyle \{U_{\alpha} \} $. From here can we say the these union of open balls is $\displaystyle X $ because of the denseness of $\displaystyle Q $? From here how do we show that the subcover is either finite or countable?