(Souslin set definition) Let $X$ be a topological space. A set $B \subseteq X$ is called Souslin set if there is a family of closed sets $\{F_s|s \in \mathbb{N}^{<\mathbb{N}} \}$ in $X$ such that $$B=\bigcup_{\sigma \in \mathbb{N^N}} \bigcap_{n=1}^{\infty} F_{\sigma\upharpoonright n}.$$
I want to prove that statement.
"If $A$ is Souslin set, then there exist a closed set $H$ in $X \times \mathbb{N^N}$ such that $A=\pi_X(H)$ where $\pi_X$ denotes the projection onto the first coordinate. "
I tried to prove it. Since $A$ is Souslin set, $A=\bigcup_{\sigma \in \mathbb{N^N}} \bigcap_{n=1}^{\infty} F_{\sigma\upharpoonright n}.$ I construct a set $H=\bigcap_{n=1}^{\infty} \bigcup_{s \upharpoonright n \in \mathbb{N}^n} F_{s \upharpoonright n} \times I(s\upharpoonright n),$ where $I(s\upharpoonright n)=\{t\in \mathbb{N^N}|t\upharpoonright n=s\upharpoonright n \}.$ I know that $\pi_X{H}=A$ but I'm not sure that $H$ is a closed set in $X \times \mathbb{N^N}.$ Is it true? How to prove that $H$ is closed in $X \times \mathbb{N^N}.$ Thank you very much.
The definition of your set $H$ is a bit confusing. Perhaps it would be clearer to write it like this: $$ H=\bigcap_{n=1}^{\infty} \bigcup_{s \in \mathbb{N}^n} F_{s} \times I(s), $$ and then $I(s)=\{t\in \mathbb{N^N}\mid t\upharpoonright n=s \}$. It is indeed true that the projection of $H$ onto the first coordinate gives you $A$.
Then it is enough now to show that each union $F:=\bigcup_{s \in \mathbb{N}^n} F_{s} \times I(s)$ is closed. One key observation is the fact that the sets $I(s)$ for $s \in \mathbb{N}^n$ form a clopen partition of $\mathbb{N^N}$. From this you will be able to conclude the proof. In any case, the details are below.