$\newcommand{\cl}{\operatorname{cl}}$We can actually prove more. Let $X$ be a $T_4$-space, and let $A$ be an $F_\sigma$-set in $X$; if $H$ and $K$ are disjoint, relatively closed subsets of $A$, then there are disjoint open sets $U$ and $V$ in $X$ (not just in $A$) such that $H\subseteq U$ and $K\subseteq V$.
One proof of this is very similar to the usual ‘climbing a chimney’ proof that a regular, Lindelöf space is normal.
Proof: There are closed sets $F_n\subseteq X$ for $n\in\Bbb N$ such that $A=\bigcup_{n\in\Bbb N}F_n$, and $F_n\subseteq F_{n+1}$ for each $n\in\Bbb N$. Note that $H\cap\cl_XK=\varnothing=K\cap\cl_XH$.
For $n\in\Bbb N$ let $H_n=H\cap F_n$ and $K_n=K\cap F_n$; the sets $H_n$ and $K_n$ are closed in $X$. (To see this, note that $H_n=H\cap F_n=(\cl_XH\cap A)\cap F_n=\cl_XH\cap(A\cap F_n)=\cl_XH\cap F_n$, which is clearly closed in $X$, and similarly for $K_n$.)
Now use the normality of $X$ to carry out the recursive construction of open sets $U_n$ and $V_n$ in $X$ for $n\in\Bbb N$ such that:
- $H_0\subseteq U_0\subseteq\cl_XU_0\subseteq X\setminus\cl_XK$;
- $K_0\subseteq V_0\subseteq\cl_XV_0\subseteq X\setminus(\cl_XH\cup\cl_XU_0)$;
- $H_{n+1}\cup\cl_XU_n\subseteq U_{n+1}\subseteq\cl_XU_{n+1}\subseteq X\setminus(\cl_XK\cup\cl_XV_n)$ for each $n\in\Bbb N$; and
- $K_{n+1}\cup\cl_XV_n\subseteq V_{n+1}\subseteq\cl_XV_{n+1}\subseteq X\setminus(\cl_XH\cup\cl_XU_{n+1})$ for each $n\in\Bbb N$.
(1) is possible because $H_0$ and $\cl_XK$ are disjoint closed sets in $X$. Then (2) is possible because $K_0$ and $\cl_XH\cup\cl_XU_0$ are disjoint closed sets in $X$: we already knew that $K_0\cap\cl_XH=\varnothing$, and by construction $K_0\cap\cl_XU_0\subseteq\cl_XK\cap\cl_XU_0=\varnothing$. The argument that (3) and (4) can be carried out is a straightforward induction.
Now let $U=\bigcup_{n\in\Bbb N}U_n$ and $V=\bigcup_{n\in\Bbb N}V_n$; clearly $U$ and $V$ are open in $X$, $H\subseteq U$, and $K\subseteq V$; I’ll leave to you the easy verification that $U\cap V=\varnothing$. $\dashv$
Your class of sets cannot be a sigma algebra because an infinite union of $G_{\delta}$ sets may not be $G_{\delta}$
Take for instance $\mathbb{Q}=\{q_1,q_2.....\}$ in the usual topology of the real line.
Then $\mathbb{Q}=\bigcup_{n=1}^{\infty}\bigcap_{i=1}^{\infty}(q_n-\frac{1}{i},q_n+ \frac{1}{i})$ a union of $G$ delta sets.
But it is proved from Baire's category theorem that $\mathbb{Q}$ is not a $G$ delta set.
But your class of sets it is an algebra though.
Lets Denote $\Sigma$ the class of such sets
Let $A \in \Sigma$
We will use the fact that in a topological space the complement of a closed set is open and the complement of an open set is closed.
$A$ is a $G_{\delta}$ set thus from De Morgan Laws $A^c$ is a $F_{\sigma}$ set.
$A$ is a $F_{\sigma}$ thus proceeding as before $A^c$ is a $G_{\delta}$ set.
So $A^c$ belongs in $\Sigma$.
Now let us take a collection of sets $A_1,A_2...A_N \in \Sigma$
$A_n$ is an $F_{\sigma}$ set for all $n \in \{1,2...N\}$ thus it is a union of closed sets i.e $A_n= \bigcup_{i=1}^{\infty}F_{n,i}$
Thus $\bigcup_{n=1}^{N} A_n=\bigcup_{n=1}^{N}\bigcup_{i=1}^{\infty}F_{n,i}$ so it is a union of closed sets.
From this we conclude that the finite union of $A_n$ is an $F_{\sigma}$ set.
.
$A_n$ is an $G_{\delta}$ set for all $n \in \{1,2....N\}$ thus it is an intersection of open sets i.e $A_n= \bigcap_{i=1}^{\infty}G_{n,i}$
Thus $\bigcup_{n=1}^{N} A_n=\bigcup_{n=1}^{N}\bigcap_{i=1}^{\infty}G_{n,i}=\bigcap_{i=1}^{\infty}\bigcup_{n=1}^{N}G_{n,i}=\bigcap_{i=1}^{\infty}B_i$
where $B_i=\bigcup_{n=1}^{N}G_{n,i}$
Note that each $B_i$ is an open set a union of open sets.
In general a finite union of $G_{\delta}$ sets is again $G_{\delta}$
From this we conlude that the union of $A_n$ is a $G_{\delta}$ set,and it is also $F_{\sigma}$
So the finite union of $A_n$ is in $\Sigma$
Now using the fact that in a topological space the whole space and the empty set are both open and closed
you can prove that $\Omega,\emptyset \in \Sigma$
Best Answer
For a metric space we can do the following: let $A$ be a non-empty subset of $(X,d)$.
Then define $f(x) = d(x,A) = \inf \{d(x,a): a \in A \}$. This is a continuous function from $X$ to $\mathbb{R}$. See this answer, e.g.
When $A$ is also closed, $A$ is exactly the set of all points with $f(x) = 0$. E.g. see this answer.
So $A = f^{-1}[\{0\}] = \cap_n f^{-1}[(-\frac{1}{n},\frac{1}{n})]$ is the countable intersection of open sets.