Can any collection of open sets in $\mathbb{R}$ be covered by a countable subcollection? Let $A$ be a collection of open sets in $\mathbb{R}$. is there a countable subcollection $G_i$ of $A$ such that $$\cup_{G\in A} G=\cup_{i=1}^\infty G_i$$
I guess there must be such subcollection, but I don't know how to establish it.

For each open interval $(a,b)$ with rational endpoints, if there is some $G\in A$ with $(a,b)\subseteq G$, then pick one such $G$.  As there are only countably many rational intervals, you'll pick only countably many $G$'s. I claim that their union equals the union of all the original $G$'s. To see this, consider any $x$ in the latter union; so $x\in G$ for some $G\in A$. As $G$ is open and $\mathbb Q$ is dense in $\mathbb R$, there are rational $a<b$ such that $x\in(a,b)\subseteq G$. But then we picked some $G'$ that includes $(a,b)$ and therefore contains $x$.
