diagram.

[c|l]{}\
\
ONE & TWO\
${\mathcal{O}}(U_{B_0})$ &\
&\
& $x_0*U_{B_0}$\
${\mathcal{O}}(U_{B_1})$ &\
&\
& $x_1*U_{B_1} $\
$\vdots$ & $\vdots$\

[c|l]{}\
\
ONE & TWO\
&\
$A_0$ & $x_0 = F(A_0)$\
&\
&\
$A_1$ & $x_1=F(A_0,A_1)$\
&\
$\vdots$ & $\vdots$\

We shall now see that $\sigma$ is a winning strategy for TWO in the game ${{\sf G}_1}({{\mathcal{O}}_{nbd}},{\mathcal{O}})$. Thus, consider a $\sigma$-play $${\mathcal{O}}(U_{B_0}),\, x_0*U_{B_0},\,{\mathcal{O}}(U_{B_1}),\, x_1*U_{B_1},\,\cdots,\, {\mathcal{O}}(U_{B_n}),\, x_n*U_{B_n},\,\cdots$$

Let $x\in G^*$ be given. We must show that $x\in\bigcup_{n<\omega}x_n*U_{B_n}$. Put $B=\cup_{n<\omega}B_n$. Then evidently $U_B\subseteq U_{B_n}$ holds for each $n$. Also $L = x*U_B$ is left coset of $U_B$ in $G^*$.

For each $n<\omega$ there is a $y\in A_n$ with $L\subseteq y*U_{B_n}$.

To see this, fix $n<\omega$. Since $U_B\subseteq U_{B_n}$ we have $x*U_B\subseteq x*U_{B_n}$. But $x*U_{B_n}$ is a left coset of $U_{B_n}$ in $G^*$, and so by the choice of $A_n$ there is a $y\in A_n$ with $x*U_{B_n} = y*U_{B_n}$.

Thus, choose for each $n$ a $y_n\in A_n$ such that $L\subseteq y_n*U_{B_n}$.

For each $n<\omega$ we have $y_{n+1}*U_{B_{n+1}}\subseteq y_n*U_{B_n}$.

For suppose on the contrary that $y_{n+1}*U_{B_{n+1}}\not\subseteq y_n*U_{B_n}$. Then we also have $y_{n+1}*U_{B_n}\not\subseteq y_n*U_{B_n}$, so that these are distinct left cosets of $U_{B_n}$ in $G^*$, and so are disjoint. But this contradicts the fact that 