Let $F_1, \dots, F_n$ be $F_{\sigma}$ sets, then there are closed sets $\{F_{1j}\}_{j=1}^{\infty}, \dots, \{F_{nj}\}_{j=1}^{\infty}$ such that $F_i = \bigcup_{j=1}^{\infty}F_{ij}$ for $i = 1, \dots, n$. Then
$$\bigcup_{i=1}^n F_i = \bigcup_{i=1}^n\bigcup_{j=1}^{\infty}F_{ij} = \bigcup_{j=1}^{\infty}\bigcup_{i=1}^nF_{ij} = \bigcup_{j=1}^{\infty}F'_j$$
where $F'_j = \bigcup_{i=1}^nF_{ij}$. As $F'_j$ is a finite union of closed sets it is closed. Therefore $\bigcup_{i=1}^n F_i$ is a countable union of closed sets, so it is an $F_{\sigma}$ set.
Likewise for $G_{\delta}$ (or take complements).
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
First, the two statements are equivalent (by taking complements). Next, note that it suffices to prove that the union of two $G_\delta$-sets is a $G_\delta$-set, as the more general result then follows by induction.
Let $F=\bigcap_{n\in\Bbb N}H_n$ and $G=\bigcap_{n\in\Bbb N}K_n$ be $G_\delta$-sets in $X$, where each $H_n$ and $K_n$ is open. Then with repeated use of the distributive laws we have
$$\begin{align*} F\cup G&=\left(\bigcap_{n\in\Bbb N}H_n\right)\cup\left(\bigcap_{n\in\Bbb N}K_n\right)\\\\ &=\bigcap_{n\in\Bbb N}\left(H_n\cup\bigcap_{k\in\Bbb N}K_k\right)\\\\ &=\bigcap_{n\in\Bbb N}\left(\bigcap_{k\in\Bbb N}(H_n\cup K_k)\right)\\\\ &=\bigcap_{\langle n,k\rangle\in\Bbb N^2}(H_n\cup K_k)\;. \end{align*}$$
Each of the sets $H_n\cup K_k$ is open, and $\Bbb N^2$ is countable, so $F\cup G$ is the intersection of countably many open sets, i.e., a $G_\delta$-set.