Prove that $\mathcal F \subsetneq \mathcal G_{\delta} \subsetneq \mathcal F_{\sigma\delta}$ I have to prove that
$\mathcal F \subsetneq \mathcal G_{\delta} \subsetneq \mathcal F_{\sigma\delta}$,
where $\mathcal F$ is the family of closed sets in $\mathbb R^{d}$,
$\mathcal G$ is the family of open sets in $\mathbb R^{d}$ and of course
$\sigma$ means countable sums, $\delta$ means countable intersections.
First inclusion, we define $cl(F)$ as the intersection of family $\{G_{n}: F\subset G_{n} \ and \ G_{n}\in \mathcal G\}$ and since $F$ is closed we get
$F = cl(F) = \bigcap_{n=1}^{\infty}G_{n}$. 
To show, that there won't be equality, just take 
$A_{1} = G, A_{2} = A_{3} =...= \mathbb R^{d}$ then we get
$\bigcap_{n=1}^{\infty}A_{n} = G\in \mathcal G$, but $G \notin \mathcal F$.
Is it correct to this moment?
EDIT:
Okey, now take a step forward what does the notation $\mathcal F_{\sigma\delta}$ means? Countable intersetions of countable sums of closed subsets or just the opposite? 
I suppose that we will use de Morgan laws. More specifically every element of $\mathcal G_{\delta}$ looks like that $\bigcap_{n=1}^{\infty}G_n = \bigcup_{n=1}^{\infty}G_{n}^c$. 
Hints would be great.
 A: I think that you have the right idea so far, but it’s not correct as it stands: you have not actually justified the assertion that every closed set in $\Bbb R^d$ is a $G_\delta$, because you have not explained how to get the open sets $G_n$. The simplest approach is probably this:

Let $F\in\mathscr{F}$. For each $n\in\Bbb Z^+$ let $$G_n=\bigcup_{x\in F}B\left(x,\frac1n\right)\;;$$ clearly $G_n$ is open, and it’s not hard to show that $F=\operatorname{cl}F=\bigcap_{n\in\Bbb Z^+}G_n\in\mathscr{G}_\delta$, so $\mathscr{F}\subseteq\mathscr{G}_\delta$. 

(If you’ve not already done so, however, you should at that point fill in the proof that $\operatorname{cl}F=\bigcap_{n\in\Bbb Z^+}G_n$.)
Your argument that $\mathscr{F}\subsetneqq\mathscr{G}_\delta$ is better, but it still needs a little minor repair: you didn’t say what $G$ is. The reader can guess that it’s an open set different from both $\varnothing$ and $\Bbb R^d$, but you should say so. It’s simplest to pick a particular set.

Let $G_1=B(\mathbf{0},1)$, where $\mathbf{0}=\langle 0,\ldots,0\rangle$ is the origin in $\Bbb R^d$, and for $n>1$ let $G_n=\Bbb R^d$; clearly $B(\mathbf{0},1)=\bigcap_{n\in\Bbb Z^+}G_n\in\mathscr{G}_\delta$, but $\langle 1,0,\ldots,0\rangle\in\big(\operatorname{cl}B(\mathbf{0},1)\big)\setminus B(\mathbf{0},1)$, so $B(\mathbf{0},1)\notin\mathscr{F}$, and $\mathscr{F}\subsetneqq\mathscr{G}_\delta$.

