(As the absentee supervisor am I allowed to pitch in?) To my mind this is least confusing if we take $A$ to be $B\cap C$ the intersection of a lightface open with a lightface closed set in the full Baire space. Here then the `game tree' for our purposes can be defined to be the set of finite partial plays in $\omega^{<\omega}$ up to some finite round. (Martin for example defines game trees which may have finite maximal branches, but let us ignore that here, and we'll only allow trees where every node can be extended.) Then the tree $T = \omega^{<\omega }\in L_{\omega_1^{ck}}$. (Similar considerations hold for trees which are subspaces of other spaces $X^\omega$ for countable $X$, and 'double-admissible' sets containing $X$ as an element, which we shall ignore.)
(If the two sets are not lightface but are say open or closed in some real parameter $z$, then the tree may be in $L_{\omega_1^{ck}}$, but as Andreas points out the game (that is the payoff set $A$) may only be defined in $L_{\omega_1^{z ck}}[z]$. But let us continue to assume the param. $z=\emptyset$.)
Now a winning strategy for the (lightface) open game $B$ on this `game-tree' is definable over (but is not necessarily a member of) $L_{\omega_1^{ck}}$ (and similarly for the closed game with payoff $C$).
For $s \in \omega^{<\omega}$ let $N_s$ be the open neighbourhood of $s$: the set of all those $\omega$-sequences extending $s$. Suppose player I has no w.s for the game playing into $A$.
Let $z$ be : $$\{ s \mid s \mbox{ is a finite even length seq. with } N_s\subseteq B\wedge
\mbox{ I has a w.s in the game with starting position } s \mbox{ and playing into } C
\}$$
This is a $\Sigma^1_1$ set and thus $z\in L_{\omega_2^{ck}}$ (but not necessarily $ L_{\omega_1^{ck}}$). The set, $D$, of $\omega$-sequences extending elements of $z$ is thus an open set in the param. $z$. By the usual argument for $\Sigma^0_1(z)$-Det. either I or II has a w.s to play into, or out of, $D$; moreover this strategy can be taken to be either in (in the case of I winning), or $\Pi_1$-definable over (in case of II), the next admissible set containing $z$: here $L_{\omega_2^{ck}}$.
Now argue that I cannot have the w.strat. in this latter game (since we assumed I had no w.s. for $G_A$). Hence II has the w.s. here. Call it $\tau$.
Let $$y = \{ s \mid s \mbox{ is a finite even length seq. with } N_s\subseteq B\wedge
s \notin z \} .$$
For each $s\in y$ let $\tau(y)$ be the $L$-least w.s for II witnessing that $s\notin z$ (using $\Sigma^0_1$-Det.) We may assume the function $ s \rightarrow \tau(s)$ with domain $y$ is $\Sigma_1(L_{\omega_2^{ck}})$ (an essential use of $\Sigma_1$-Rep. here.) Now define a strategy $\tau^*$ by setting $\tau^*(s) =\tau(s)$ if $N_s\not\subseteq B$ and $\tau^*(s\smallfrown r) = \tau(s)(r)$ where $N_s \subseteq B$ (and $s$ is the shortest init. set. of $s \smallfrown r$ for which the latter is true). Now argue that $\tau^*$ is a w.s. for II in $G_A$.
This shows that Det$(\Sigma^0_1 \cap \Pi^0_1)$ holds definably over (rather than strictly within) the second admissible set over the tree.)
Lastly we point out that strategies for such games cannot all be elements of $L_{\omega_2^{ck}}$ - or in other words of $HYP(\mathcal{O})$: the complete $\Game(\Sigma^0_1 \cap \Pi^0_1)$ set of integers will be $\Pi_1(L_{\omega_2^{ck}})$.
