$\emptyset\neq L\subseteq y_{n+1}*U_{B_{n+1}}\cap y_n*U_{B_{n}}$.

Towards proving the next claim first note that if $x*U_{B_n}\subseteq y*U_{B_n}$ then $(\forall \alpha\in B_n)(x(\alpha)=y(\alpha))$.

For each $n<\omega$ we have $support(y_n)\subseteq support(y_{n+1})$.

For by Claim 2 we find a $u\in U_{B_n}$ such that $y_{n+1} = y_n*u$. Now $support(y_n)\cap support(u) = \emptyset$, and so $support(y_{n+1}) = support(y_n)\cup support(u)\supseteq support(y_n)$.

For each $n<\omega$ we have $support(y_n)\subseteq support(x)$.

Since $x\in L\subseteq y_n*U_{B_n}$ it follows that $y_n^{-1}*x\in U_{B_n}$, and so for each $\alpha\in B_n$ we have $x(\alpha) = y_n(\alpha)$. Since $support(y_n)\subseteq B_n$, Claim 4 follows.

Since $x$ is in $G^*$ it has finite support. Claims 3 and 4 imply that there is a $k$, from now on fixed, such that for all $n\geq k$ we have $support(y_n) = support(y_k)$. It follows that for all $n\ge k$, $y_n = y_k$. But then, for all $n\ge k$, $x\in y_k*U_{B_n}$, which implies that $x\in y_k*U_B$. But for infinitely many $n$ we have $x_n = y_k$,a nd so for such an $n$ larger than $k$, $x\in x_n*U_{B_n}$.

This completes the proof that $\sigma$ is a winning strategy for TWO. $\Box$

From this we now derive that TWO in fact has a winning strategy in the game ${{\sf G}_1}({\mathcal{O}},{\mathcal{O}})$, typically a harder game for player 