
open cover
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?

I'm not very good in English, but I don't think $\displaystyle Q$ is a cover...
I think you mean X is separable, and Q is countable and dense in X... I don't know if there is a english name for this kind of set...
First of all, you should note $\displaystyle \mathcal{B}$ is countable:
$\displaystyle \mathcal{B} = Q \times \mathbb{Q} = \aleph_0 \aleph_0 = \aleph_0$
Secondly $\displaystyle E$ is not a subcover of $\displaystyle \{U_{\alpha}: \alpha \in A \}$.
Thirdly, the idea is good, and $\displaystyle E$ covers X:
Let $\displaystyle x \in X$. Then $\displaystyle x \in U_\alpha$ for some $\displaystyle \alpha$. $\displaystyle U_\alpha$ is neighbourhood hence there is the ball $\displaystyle B_{r}(x) \subset U_\alpha$. Q is dense in X so there is $\displaystyle q \in Q$ such that $\displaystyle d(x, q) < r/3$. There exists ball $\displaystyle B_\epsilon(q) \in \mathcal{B}$ where $\displaystyle r/3< \epsilon < r/2 $. Thus $\displaystyle x \in B_\epsilon(q)$, and $\displaystyle B_\epsilon(q) \subset B_r(x)$.
Of course $\displaystyle B_\epsilon(q) \in E$. Hence $\displaystyle E$ (and $\displaystyle \mathcal{B}$ too) is cover of X.