1. ## 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?

2. 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.