Uncountable analytic sets contain compact Cantor copies

Fix continuous $f:\mathcal{N}\to X$ onto A by F1 and A1. We build words $t_s\in {\mathbb{N}}^{<\omega }$ indexed by binary words s, with $t_{\mathrm{\varnothing }}=\mathrm{\varnothing }$, strict extension on each edge, and uncountable $f[N_{t_s}]$. Given $t_s$, F2 supplies disjoint opens $U_0,U_1$ meeting its image uncountably. Each $N_{t_s}\cap f^{-1}[U_i]$ is open and is the union of all cylinders it contains whose word lengths exceed $\mathrm{\mid }{t}_s\mathrm{\mid }$. There are countably many such cylinders. If every one had countable image, A1 would choose enumerations of the nonempty images and a pairing would enumerate their union, contradicting its uncountability. Thus for each i select the least word code with uncountable image and cylinder inside this preimage, and set it to $t_{s^{⌢}i}$. It extends $t_s$ and has its image inside $U_i$. All choices of eligible open pairs can be made on the set of finite words by A1; length recursion then constructs the tree of words.

For $z\in \mathcal{C}$ set $g(z)={\bigcup }_n{t}_{z\upharpoonright n}$. Strict length growth makes this a full Baire sequence. Agreement on n input bits fixes an output prefix of length at least n, so g is continuous by F3. If z,w first split at a binary node s, their images under $f\circ g$ lie in the two disjoint opens chosen there in step 1.1. Hence $f\circ g$ is injective as well as continuous, and its image is contained in A.

The image K is compact by pulling any open cover back to compact $\mathcal{C}$ and pushing a finite subcover forward. For a point x outside a compact subset of a metric space, the balls $B(y,d(x,y)\mathrm{/}3)$ about its members y have a finite subcover; the minimum of these finitely many positive radii gives a ball about x missing the compact set. Hence compact sets are closed. Closed subsets of $\mathcal{C}$ are compact (adjoin the open complement to a cover), so their images under $f\circ g$ are closed. The inverse of this injection onto K is therefore continuous. Thus K is homeomorphic to $\mathcal{C}$, is nonempty and closed, and has no isolated point by F3. Finally Borel A is analytic by F1's normal forms (use its identity map), and X is itself Borel in X. This proves both final special cases. QED.