Analytic countable operations and inclusion of Borel sets

By F1 and A1 choose closed $F_n\subseteq X\times \mathcal{N}$ projecting to $A_n$. For $z\in \mathcal{N}$ write $z^{-}=(z(1),z(2),\dots )$ and put $F=\{(x,z):(x,z^{-})\in F_{z(0)}\}$. This is closed: at a point outside it, fixing $z(0)=n$ and a product neighbourhood outside closed $F_n$ gives a neighbourhood outside $F$. Its projection is ${\bigcup }_n{A}_n$: a witness $z$ gives the summand $n=z(0)$; a witness $y$ in a summand gives $z=n^{⌢}y$. Hence the union is analytic.

If $g:Y\to X$ is continuous between Polish spaces and $A$ has closed witness $F$, the map $(y,z)\mapsto (g(y),z)$ is continuous: product basic opens pull back to products of open preimages and cylinder opens. Its preimage of $F$ is closed and projects exactly to $g^{-1}[A]$. All product spaces here are Polish by F3, so F1 applies.

Decode $z$ as $(z_n{)}_n$ by F2 and put $H=\{(x,z):\mathrm{\forall }n(x,z_n)\in F_n\}$. Each coordinate map is continuous, so its closed preimage is closed by F5 and complementation; their intersection is closed. A projected point is in every $A_n$. Conversely if $x\in {\bigcap }_n{A}_n$, A1 selects witnesses $y_n$ from the nonempty closed sections $(F_n{)}_x$, and F2 codes them as $z$ with $(x,z)\in H$. Thus this projection is exactly the intersection, analytic by F1.

A closed $C\subseteq X$ has the closed witness $C\times \mathcal{N}$; this projects to $C$ since the all-zero Baire real exists. By F4 and step 1.1 every open is analytic. Let $\mathcal{H}$ consist of sets both they and their complements analytic. It contains opens and is complement closed. If $A_n\in \mathcal{H}$, step 1.1 makes their union analytic and step 2.1 makes its complement, the intersection of their analytic complements, analytic. Thus $\mathcal{H}$ is a sigma-algebra containing opens and contains every Borel set by the leastness definition. Empty intersections give $X$, and empty unions give $\mathrm{\varnothing }$, both already supplied by the closed-witness construction. QED.