Closed Souslin schemes characterize analytic sets

For a closed scheme $F$ put $C=\{(x,f):\mathrm{\forall }nx\in F_{f\upharpoonright n}\}$. If $(x,f)\mathrm{\notin }C$, some n has $x\mathrm{\notin }{F}_{f\upharpoonright n}$. The open product $(X\setminus F_{f\upharpoonright n})\times N_{f\upharpoonright n}$ misses $C$. This remains true for n=0. Hence $C$ is closed and its projection, exactly $\mathsf{S}(F)$ by F1, is analytic by F2 and A1.

Conversely empty $A$ uses the all-empty closed scheme. If $A$ is nonempty analytic, F2 and A1 give continuous $g:\mathcal{N}\to X$ with image $A$. Set $F_s=\overline{g[N_s]}$, closed and decreasing. For each $f$, $g(f)$ belongs to all $F_{f\upharpoonright n}$. If $x\mathrm{\ne }g(f)$, put $r=d(x,g(f))\mathrm{/}3>0$. Continuity gives n with $g[N_{f\upharpoonright n}]\subseteq B(g(f),r)$; its closure lies in the closed radius-r ball, which excludes x. Thus ${\bigcap }_n{F}_{f\upharpoonright n}=\{g(f)\}$. Taking the branch union gives exactly $A$. The closures, rather than the raw images, supply closed sets without changing the branch intersections. QED.