AD implies the Baire property for subsets of sequence spaces and the real line

Let U be the union of the basis opens V on which A is comeagre. For each contributing V, F1's open-subspace transfer gives a sequence of ambient closed nowhere dense sets covering $V\setminus A$. Such a sequence has a Baire-real code: each closed F is determined by the set of basis indices whose opens miss F, since their union is exactly $X\setminus F$. Code this binary index set, and pair its coordinates with the sequence index to code the whole sequence in ${\mathbb{N}}^{\mathbb{N}}$. For each contributing V the set of valid covering codes is nonempty; for every other V take the singleton code of the all-empty sequence. F2 under the assumed AD selects one code per basis index. Decode and pair the two sequence indices. This gives an actual closed nowhere dense covering sequence for $U\setminus A$.

Put $E=A\setminus U$. AD determines its coded category game. If I won, F1 would make E comeagre in some nonempty basic V. Since $E\subseteq A$, the same witnesses show A comeagre in V; hence $V\subseteq U$ by definition of U. But then $E\cap V=\mathrm{\varnothing }$, so the same witnesses would make V meagre in itself, contradicting F1. Thus I cannot win, and determinacy gives a winning II strategy. F1 provides a specified nowhere dense covering sequence for E.

Interleave that sequence with step 1.1's sequence for $U\setminus A$. Their union covers $A\mathrm{△}U=E\cup (U\setminus A)$ and each term is nowhere dense. Hence the symmetric difference is meagre; U is open by its definition, so F3 proves the Baire property. This construction works with rational interval codes for the real line as well as with the two cylinder bases; it requires no homeomorphic transfer. Empty A gives U empty and the same argument, while A=X gives U=X. QED.