AD gives the perfect-set property in sequence spaces and the real line

For $A\subseteq \mathcal{C}$, F1's explicit finite-block codes and illegal-bit payoff make its game a natural-number game, determined by A1. If I wins, F1 gives a compact Cantor copy in A; if II wins, F1 gives an injection $A\to \mathbb{N}$. These cover every case, including empty A.

For $A\subseteq \mathcal{N}$, apply step 1.1 to h[A]. If it injects into $\mathbb{N}$, compose that injection with h restricted to A. If it contains a compact Cantor copy K, F2's continuous inverse on D restricts to K, giving a homeomorphic copy in A that is compact by the open-cover definition. In a metric space compact sets are closed: for an exterior x finitely many balls $B(y,d(x,y)\mathrm{/}3)$ cover the compact set, leaving a sufficiently small ball about x disjoint. Homeomorphism with Cantor space gives no isolated point by F2. Thus this is also a nonempty perfect closed subset of Baire space.

For $A\subseteq \mathbb{R}$, enumerate integers as $m_i=0,1,-1,2,-2,\dots$ and set $A_i=(A\cap [m_i,m_i+1))-m_i$. By F5 these pieces cover A after translation. Apply step 1.1 to each b[A_i]. If one contains a compact Cantor copy, F3's ZF clause transfers it into A_i, and translation, with its continuous inverse, transfers it into A. It is compact and therefore closed by the separation argument in step 2.1, and has no isolated point.

Otherwise each b[A_i] injects into $\mathbb{N}$. Each nonempty such set has a surjective enumeration: invert an injection on its range and fill unused indices with the value at the least occupied index. A sequence of binary reals is coded as one Baire real by F2's pairing. For each nonempty b[A_i] let C_i be the nonempty set of all codes of its enumerations; for an empty piece use the singleton all-zero code and retain that it was empty. F4 with A1 selects c_i for every i. Decode them, apply F3's $\pi$, translate back by m_i and interleave the two indices. If A is nonempty, fill slots from empty pieces with one fixed element a of A. This gives a surjection $\mathbb{N}\to A$ because every point belongs to a piece and appears in that piece's enumeration. Assign each point its least enumeration index to inject A into $\mathbb{N}$. For empty A use the empty injection. No unrestricted countable choice has been used. QED.