Countable Union in separable metric space

Question

I am Reading a book on measure theory ("Probability measures on metric spaces", Phartasarathy) and I fuond a sentence which is not proven:
Let S be a separable metric space. Let $\mathcal B$ a collection of open sets $B_\alpha$. Then there exists a numerable sub-collection $(B_n)_n\in \mathcal B$ such that
$$\bigcup_\alpha B_\alpha=\bigcup_n B_n$$
In other words: every union of open sets can be written as a countable union of open sets belonging to the previous ones.
It is more about topology than about measure theory and I have no idea how to prove It.

Answer 1

A separable metric space is second countable. So let $\{U_n\}_{n=1}^\infty$ be a countable basis of $S$. Now let:

$M=\{n\in\mathbb{N}: \exists\alpha\ \ U_n\subseteq B_{\alpha}\}$

For each $n\in M$ we can choose $\alpha_n$ such that $U_n\subseteq B_{\alpha_n}$. Now we can prove that $\cup_{\alpha} B_{\alpha}=\cup_{n=1}^\infty B_{\alpha_n}$. Let's say $x$ belongs to the union on the left side. Then there is $\alpha$ such that $x\in B_{\alpha}$. But from one of the definitions of a basis we know that there is $k\in\mathbb{N}$ such that $x\in U_k\subseteq B_{\alpha}$. This implies $k\in M$ and then $x\in U_k\subseteq B_{\alpha_k}$.