Let $\mathbb N^{\mathbb N}=\mathscr N,$ and $A$ a subset of the Polish space, $X.\ A$ is said to be analytic if it is the continuous image of a function from a Polish space. A standard result is that a subset $A$ of $X$ is analytic if and only if there is a closed subset of $\mathscr N ×X$ whose projection on $X$ is $A.$ We also have that if $X$ is Polish, then the closed sets in $X$ may be indexed by $\mathscr N,$ which is to say there is a closed subset $A$ of $\mathscr N\times X $ such that the closed sets in $X$ are precisely the sections $A_{\bf n}=\left \{ x\in X:(\textbf{n},x)\in A \right \}$
The exercise is to show that there exists an analytic set that is not Borel, and the hint is to consider the projection $\mathscr N\times \mathscr N\times X\to \mathscr N\times X;\ (\textbf{m},\textbf{n},x)\to (\textbf{m},x)$, to show that the analytic sets in $X$ may also be indexed by $\mathscr N$.
Here is my attempt, with the questions in italics:
To start, note that there is a closed subset $S$, of $\mathscr N\times \mathscr N\times X$ such that $\left \{ S_{\textbf{n}} \right \}_{\textbf{n}}$ constitutes the closed sets of $\mathscr N\times X.$ Furthermore, the map $\pi:(\textbf{m},\textbf{n},x)\mapsto (\textbf{m},x)$ is continuous so its image is analytic because $\mathscr N\times \mathscr N\times X$ is Polish. Let $A$ be the image of this map and consider the sections $A_{\bf m}.$ We can define the projection $\pi_1:S_{\textbf{m}}\to A_{\bf m}$ by $(\textbf{n},x)\mapsto x$ so that each $A_{\textbf{m}}$ is the projection from $\mathscr N\times X$ of the closed set $S_{\textbf{m}}$ and therefore is analytic. On the other hand, if $A'$ is analytic, then it is the projection of some closed set in $\mathscr N\times X$, which in turn, must be one of the $S_{\bf m}.$ This proves the claim in the hint.
Is the foregoing rigorous enough? If not, can you suggest how to make it so?
From here, it's a standard argument but I ran into a snag:
There is an analytic subset $S$ of $\mathscr N\times \mathscr N$ such that the analytic sets in $\mathscr N$ are precisely the sections $S_{\textbf{n}}.$ Now define $D=\left \{ \textbf {n}:(\textbf {n},\textbf {n})\in S \right \}.$
Now I want to claim that $D$ is analytic because it is the projection on $\mathscr N$ of the closed set $S\cap \left \{ (\textbf n,\textbf m):m=n \right \}.$
But I can't show this set is closed.
edit: the set is not necessarily closed but it is analytic, being the intersection of two analytic sets. So the proof proceeds.
If $D$ is Borel, then so is $\mathscr N\setminus D$, which, by a routine argument, is seen to be analytic. So there must be an $\textbf i$ for which $S_{\textbf i}=\mathscr N\setminus D.$
Now, if $\textbf i\in D$ then $\textbf i\in S_{\textbf i},$ which is a contradiction. Symmetry shows that if we assume that $\textbf i\in \mathscr N\setminus D,$ we reach impossible conclusion that $\textbf i\in D.$
And this finishes the proof.