Suppose we have a compact metric space $X$ with some designated point $a$ and closed set $B$ such that $a\notin B$. Let $A_0=\{a\}$ and $B_0 = B$. We'll play a game where on each player's turn they choose closed sets $A_{n+1}$ and $B_{n+1}$ with the requirements that $A_n\subseteq A_{n+1}^\circ$ and $B_n\subseteq B_{n+1}^\circ$ and $A_{n+1}\cap B_{n+1}=\varnothing$. After $\omega$ turns, let $A_\omega=\bigcup_{n<\omega}A_n$ and $B_\omega=\bigcup_{n<\omega}B_n$, which are open sets, and let $C=X-(A_\omega \cup B_\omega)$.
The broad question is: What properties can I force $A_\omega$, $B_\omega$, and $C$ to have regardless of what moves you play? For instance: I'm pretty sure that I can force that $A_\omega$ and $B_\omega$ are regular open sets and $C$ is both of their boundaries: On turn $n$ I can make it so that the no point in $X-(A_n \cup B_n)$ has distance more than $2^{-n}$ from $A_n$ and distance no more than $2^{-n}$ from $B_n$, which in the limit ensures that every point in $C$ is a limit point of both $A_\omega$ and $B_\omega$ individually. I'm also pretty sure that if the topological dimension (all the common notions are equivalent for separable metric spaces) of $X$ is $\leq n$, then I can force $\text{dim }C\leq n-1$.
I'm interested in that question in general, but I'm also curious about a particular case: If $X=[0,1]^2$, $a\in(0,1)^2$, and $[0,1]^2 - (0,1)^2\subseteq B$, can I force it that $C$ has a closed subset $D$ which is a Jordan curve clopen in $C$ (i.e. $C-D$ is also closed) surrounding $a$. I would guess the answer is no, but I have no idea how to prove it one way or the other.
