A possible easy proof of Borel determinacy? I was initially trying to prove determinacy of $\mathbf{\Sigma}^0_2$ games, but surprisingly the proof I came up with seems to generalise all the way to Borel determinacy. Obviously this sounds a little too good to be true, but it still appears correct to me after double-checking many times. Perhaps someone can help me spot mistakes which I might have overlooked?
The proof:
Lemma. Suppose all $\mathbf{\Pi}^0_\alpha$ games are determined. Let $G(T,A)$ be such a game, and suppose $II$ has a winning strategy. If $W_{II}$ is $II$'s canonical quasistrategy, and $s = \langle x_1,\ldots,x_{2k}\rangle$ is a branch such that $\langle x_1,\ldots,x_{2k-1}\rangle\in W_{II}$ but $s\notin W_{II}$, then $I$ has a winning strategy in the game $G(T_s,A\cap N_s)$.
Proof. The subgame $G(T_s,A\cap N_s)$ is also $\mathbf{\Pi}^0_\alpha$ and hence determined. So if $I$ doesn't have a winning strategy, then $II$ does. But then this implies that the $W_{II}$ can be extended to a larger (winning) quasistrategy containing the winning strategy from the branch $s$, and we have $s\in W_{II}$ which is a contradiction. $\square$
Let $G=G(T,A)$ be a $\mathbf{\Sigma}^0_\alpha$ game, where $A=\bigcup_n A_n$ and each $A_n$ is $\mathbf{\Pi}^0_{\gamma_n}$ with $\gamma_n<\alpha$. By induction, the games $G_n=G(T,A_n)$ are determined. If $I$ has a winning strategy for at least one of the $G_n$'s, then this is also a winning strategy for $G$.
Otherwise, $II$ has a canonical quasistrategy $(W_{II})_n$ for each $G_n$. Let $W$ be the intersection of these quasistrategies (when viewed as sets of branches).
Case 1. If $W$ contains a strategy, then this is automatically a winning strategy for $II$ in $G$.
Case 2. Otherwise, $W$ contains what I call conflicting branches: a branch $s=\langle x_1,\ldots,x_{2k-1}\rangle$ is conflicting if

*

*$s\in W$.

*For all $x_{2k}$ satisfying $s\frown x_{2k}\in T$, there is some $n$ such that $s\frown x_{2k} \notin (W_{II})_n$.

If $s$ is conflicting, then regardless of $II$'s next move $x_{2k}$, by the lemma $I$ always has a winning strategy in some $G_n$ (hence $G$) starting from $s\frown x_{2k}$. Furthermore, $I$ has a strategy that always leads to some conflicting branch (otherwise $II$'s counter-strategy that avoids all conflicting branches is itself a substrategy of $W$, which is a contradiction). Therefore, $I$ has a winning strategy in $G$ overall.
Lastly, determinacy of $\mathbf{\Sigma}^0_\alpha$ games implies determinacy of $\mathbf{\Pi}^0_\alpha$ games, completing the inductive step.
 A: This has been answered in the comments; I'm posting to move this off the unanswered queue, and have made this CW to avoid reputation gain.

The problem with your argument, as Jonathan observed above, is that the "canonical quasistrategies" are actually extremely weak objects. Take for example the game on $\{0,1\}$ where player $II$ wins as long as they ever play a "$1$." Obviously every state is a winning state for player $II$, so the canonical quasistrategy in this case is ... "do whatever you want." But there are clearly plays consistent with this quasistrategy according to which player $II$ loses.
Of course the above nonsense can't happen if the payoff set is closed - this is why closed determinacy is so easy to prove! This indicates a purely topological difficulty in proving determinacy: the set of plays consistent with a given quasistrategy will always be closed, so essentially we need all of our winnable payoff sets to have perfect subsets. Note that it's already not obvious that every Borel set has a perfect subset (or has complement with a perfect subset); this can be proved using closed determinacy + unfolding (see Kechris' book), but is definitely nontrivial.
The fix, very broadly speaking, is to look at quasistrategies with promises, which prevent the sort of "silly-loss-by-infinite-delay" - as well as its less-silly versions - that cropped up above. This ultimately takes the form of converting an arbitrary rank Borel game on $\mathbb{N}$ or similar into a closed game on a much more complicated set. Interestingly, this is a bit more than we need for determinacy - it's a stronger property called "unravelability," whose exact extent seems much more mysterious.
