I have a question concerning the proof of this theorem in Kechris' "Classical Descriptive Set Theory." The rules of the game i as follows: Let $X$ be a nonempty perfect Polish space with compatible complete metric $d$. Fix a basis $\{V_n\}$ of nonempty open sets for $X$. Given $A\subseteq X$, consider the game $G^*(A)$, where Player 1 (I) plays pairs of basic open sets (with certain requirements, but they are not necessary for my question), while Player 2 (II) moves by choosing one of those open sets, I plays two open sets from the set that II chose, and so on... (p. 149, Kechris).
Now, in the proof of $\Rightarrow$, Kechris does as follows: Assume $\sigma$ is a winning strategy for II. Given $x\in A$, call a position \begin{align*} p=((U_0^{(0)},U_1^{(0)}), i_0,...,(U_0^{(n-1)},U_1^{(n-1)}),i_{n-1}), \end{align*} $i_j\in \{0,1\}$, "good" for $x$, if it has been played according to $\sigma$ and $x\in U_{i_{n-1}}^{(n-1)}$. He then concludes that for each $x\in A$ there is a maximal good $p$ for $x$. Now, if $p$ is maximal good for $x$, then \begin{align*} x\in A_p =\{y\in U_{i_{n-1}}^{(n-1)}:\forall \,\, \text{legal} \,\, (U_0^{(n)},U_1^{(n)}), \,\, \text{if i is what}\,\, \sigma\,\, \text{requires II to play next, then}\,\, y\not \in U_i^{(n)}\}. \end{align*} He then claims that $A_p$ contains at most one point, and this is what I cannot convince myself of! Can anyone help me? :)