Every open subset of $\mathbb R$ is an $F_\sigma$-set and a $G_\delta$-set. 
Prove that every open subset of $\mathbb R$ is an $F_\sigma$-set and a $G_\delta$-set.

In order to prove this, the exercise tells me I need to make use of the following previously proven facts:

  
*
  
*Every open interval in $\mathbb R$ is both $F_\sigma$-set and $G_\delta$-set.
  
*$\mathcal B = \{ (a,b) \in \mathbb R: a<b~ \text{and}~ a,b, \in \mathbb Q \}$ is a basis for the Euclidean topology on $\mathbb R$.
  

I have done the following so far:
We have proven that every open interval $(a,b) \subset \mathbb R$ is both a $F_\sigma$-set and $G_\delta$-set. We now attempt to prove that every open subset $S \subset \mathbb R$ is also both a $F_\sigma$-set and $G_\delta$-set. 
Since $S$ is open in $\mathbb R$, we know that it is a union of open intervals, that is 
    \begin{align*}
 S &= \bigcup_{j \in J} (a_j, b_j) \\
 &= \bigcup_{j \in J} \left[ \bigcup_{n=1}^\infty \left[ a_j + \frac{1}{n} , b_j - \frac{1}{n}\right]\right],
  \end{align*} which is a countable union of closed sets. That is, $S$ is an $F_\sigma$-set.
I now need help proving that $S$ is also a $G_\delta$-set. I am asssuming this will make use of (2) that they told us to use. I cannot seem to see how to do this though?
 A: It is trival. Since 
$$
S=\bigcap_{n=1}^{\infty}S \quad\text{where }\quad S=\bigcup_{j \in J} (a_j, b_j) $$
S is $G_{\delta}$ for $S$ is open.
A: Let $U$ be open in $\mathbb R$. Then $U$ is a countable union of disjoint intervals, so we may as well assume $U$ itself is an interal, $(a,b)$. 
If both $a$ and $b$ are finite then take $F_n=[a+1/n,b-1/n]$. Clearly $\bigcup_{n\in \mathbb N} F_n=U$ and we are done.
If $a=-\infty$. Then take $F_n=[-n,b-1/n]$ and again, $\bigcup_{n\in \mathbb N} F_n=U$.
The other two cases are similar.
Remark: You do not need to use the fact that $U$ is a countable union of intervals; the result is true in any metric space $(X,d)$. This can be seen by considering $F_n=\{x\in U:d(x,X\setminus U)\ge 2^{-n}\}$.
A: You have to do slightly better. You have to show that every open set is a $countable$ union of disjoint open intervals.
Lemma 0 Every open subset of $\mathbb{R}$ is a countable disjoint union of open intervals, and is hence measurable.
First seperate the open subset $U \subset \mathbb{R}$ into components by imposing the equivalence relation $x~y$ iff $(x,y) \in U$. Use this to prove that each component is either an interval or a ray.
hint: define $a= \inf(U)$, $b=\sup(U)$. Let $x \in (a,b)$. Then find an open ball about $x$ so that it is between $a$ and $b$. The result will follow shortly thereafter.
Then utilize the countable basis for $\mathbb{R}$ defined by
$$\mathbb{B}= \{\beta(q, 1/n) \mid q \in \mathbb{Q}, n \in \mathbb{N}\}$$, and use the fact that the rationals are dense in $\mathbb{R}$, and hence there are only countably many disjoint components.
I've given this same hint in a previous answer here
You can then proceed in the way you have done to show that each open interval is a countable union of closed sets to obtain that each open set is $F_{\sigma}$.
$G_{\delta}$ is easier, since it is the countable intersection of open sets. Well, let $U \subseteq \mathbb{R}$ be open. Define $U_{n}=U$ for all $n \in \mathbb{N}$. Then $\bigcap_{n \in \mathbb{N}} U_n=U$ and is thus open. So every open set is $G_{\delta}$
