why is each open set an $F_\sigma$?

In Royden's Real Analysis:

On $\mathbb{R}$, a set which is a countable union of closed sets is called an $F_\sigma$. Thus every countable set is an $F_\sigma$, as is, of course, every closed set. A countable union of sets in Fa is again in $F_\sigma$. Since $$(a, b)= \cup_{n=1}^\infty [a + 1/n, b - 1/n],$$ each open interval is an $F_\sigma$, and hence each open set is an $F_\sigma$.

Each open interval is an $F_\sigma$, but why is each open set an $F_\sigma$? An open set can be uncountable union of open intervals.

If we are considering a general Borel sigma algebra instead of $B(\mathbb R)$, will each open set be an $F_\sigma$?

Thanks!

Every open set in $\Bbb R$ can be written as a union of open intervals with rational endpoints, and there are only countably many such open intervals. Each of them is an $F_\sigma$, so every open set is an $F_\sigma$.

More generally, let $\langle X,d\rangle$ be a metric space, and let $U\subseteq X$ be open and non-empty. For $n\in\Bbb N$ let $F_n=\{x\in U:d(x,X\setminus U)\ge 2^{-n}\}$; each $F_n$ is closed, and $U=\bigcup_{n\in\Bbb N}F_n$. Thus, every open set in a metric space is an $F_\sigma$.

Now let $X$ be the space $\omega_1$ with the order topology. The set $U$ of isolated points of $X$ is open and uncountable. $X$ is countably compact, however, so the only closed subsets of $U$ are the finite subsets, and $U$ cannot be an $F_\sigma$.

As a slight expansion of Brian Scott's argument every open set in $\mathbb R$ is the disjoint union of a countable number of open intervals. In particular if $U$ is an open subset of $\mathbb R$ then for each $x \in U$ we can consider the largest interval $x \in I_x \subset U$. Then the set $\mathcal I=\{I_x : x \in U\}$ is a disjoint open cover of $U$ whose union is precisely $U$. We see that it's countable from observing that each interval contains a rational so there's an injection from $\mathcal I$ to the rationals.