As JDH observed, there are topological spaces $X$ for which Player II has a winning strategy (and in fact a winning 2-tactic), but no winning tactic in the topological Banach-Mazur game. So we proceed by attempting to find a poset that mimicks the topological game.

Naively one might let choose $\mathcal T_X\setminus\{\emptyset\}$. But as observed by Will Brian, this isn't quite right: in the topological game, P2 doesn't win by finding an open set below all chosen sets; they win if these sets have non-empty (not necessarily open) intersection.

Another attempt might be to "add in" a minimal point at the bottom of a chain of open sets with non-empty but empty-interior intersection to mimic this win condition for P2. But then P2 would just pick this new valid move immediately and end the game.

So instead, for each such chain in $\mathcal T_X\setminus\{\emptyset\}$, take each open set $U$ that contains the non-empty intersection, and insert a copy of $\mathcal T_U\setminus\{\emptyset\}$ below the chain. Rinse and repeat for these copies (taking only open subsets of the top $U$ for the copy), $\omega_1$ times. At the limits, the same thing is happening: any descending chain still corresponds to a bunch of open sets, so if their intersection is nonempty, add appropriate copies.

I think it's clear that if the players agree to play within the top $\mathcal T_X\setminus\{\emptyset\}$, the result is the same as the topological game. So we should argue that each player can enforce equivalent play to show strategies in the topological game translate here.

To see this, note that given however many moves of the opponent they're allowed to see, a player with a winning topological strategy/tactic can obtain an open set $U$ and choose the top element of a copy of $\mathcal T_U\setminus\{\emptyset\}$. (If they couldn't, then all infinite descending chains have empty intersection and the game is already set for P1.) The opponent then has to choose a subset of $U$ for the rest of the game.

So the winning player has ensured the game descends down our poset copies countably-far, and both players are essentially picking open subsets of each other. Let $\alpha<\omega_1$ be the least limit ordinal below this descent. If the open subsets have empty intersection, we didn't insert anything below that chain; otherwise, we did. This preserves the winning strategy/tactic for whichever player had one.

It remains to be shown that winning strategies in the poset game produce winning strategies in the topological game. Given a winninng strategy for either player in the poset game, note that it must handle the situation where the opponent always chooses the top of a lower copy of $\mathcal T_U$ for each move. This enforces the winning strategy to essentially choose subsets of the opponent at each step. Then this can be translated down to the topological game. If the poset game resulted in no lower bound, it's because the intersection of open sets was empty; otherwise it's because the intersection of open sets was non-empty.

Edit: leaving this up just in case (and for the comments), but I'm pessimistic that the idea is recoverable - see discussion below.

As JDH observed, there are topological spaces $X$ for which Player II has a winning strategy (and in fact a winning 2-tactic), but no winning tactic in the topological Banach-Mazur game. So we proceed by attempting to find a poset that mimicks the topological game.

Naively one might let choose $\mathcal T_X\setminus\{\emptyset\}$. But as observed by Will Brian, this isn't quite right: in the topological game, P2 doesn't win by finding an open set below all chosen sets; they win if these sets have non-empty (not necessarily open) intersection.

Another attempt might be to "add in" a minimal point at the bottom of a chain of open sets with non-empty but empty-interior intersection to mimic this win condition for P2. But then P2 would just pick this new valid move immediately and end the game.

So instead, for each such chain in $\mathcal T_X\setminus\{\emptyset\}$, take each open set $U$ that contains the non-empty intersection, and insert a copy of $\mathcal T_U\setminus\{\emptyset\}$ below the chain. Rinse and repeat for these copies (taking only open subsets of the top $U$ for the copy), $\omega_1$ times. At the limits, the same thing is happening: any descending chain still corresponds to a bunch of open sets, so if their intersection is nonempty, add appropriate copies.

I think it's clear that if the players agree to play within the top $\mathcal T_X\setminus\{\emptyset\}$, the result is the same as the topological game. So we should argue that each player can enforce equivalent play to show strategies in the topological game translate here.

To see this, note that given however many moves of the opponent they're allowed to see, a player with a winning topological strategy/tactic can obtain an open set $U$ and choose the top element of a copy of $\mathcal T_U\setminus\{\emptyset\}$. (If they couldn't, then all infinite descending chains have empty intersection and the game is already set for P1.) The opponent then has to choose a subset of $U$ for the rest of the game.

So the winning player has ensured the game descends down our poset copies countably-far, and both players are essentially picking open subsets of each other. Let $\alpha<\omega_1$ be the least limit ordinal below this descent. If the open subsets have empty intersection, we didn't insert anything below that chain; otherwise, we did. This preserves the winning strategy/tactic for whichever player had one.

It remains to be shown that winning strategies in the poset game produce winning strategies in the topological game. Given a winninng strategy for either player in the poset game, note that it must handle the situation where the opponent always chooses the top of a lower copy of $\mathcal T_U$ for each move. This enforces the winning strategy to essentially choose subsets of the opponent at each step. Then this can be translated down to the topological game. If the poset game resulted in no lower bound, it's because the intersection of open sets was empty; otherwise it's because the intersection of open sets was non-empty.

As JDH observed, there are topological spaces $X$ for which Player II has a winning strategy (and in fact a winning 2-tactic), but no winning tactic in the topological Banach-Mazur game. So we proceed by attempting to find a poset that mimicks the topological game. (It's late and the post has gotten long, so I'm submitting what I believe is a partial answer for now.)

Naively one might let choose $\mathcal T_X\setminus\{\emptyset\}$. But as observed by Will Brian, this isn't quite right: in the topological game, P2 doesn't win by finding an open set below all chosen sets; they win if these sets have non-empty (not necessarily open) intersection.

Another attempt might be to "add in" a minimal point at the bottom of a chain of open sets with non-empty but empty-interior intersection to mimic this win condition for P2. But then P2 would just pick this new valid move immediately and end the game.

So instead, for each such chain in $\mathcal T_X\setminus\{\emptyset\}$, take each open set $U$ that contains the non-empty intersection, and insert a copy of $\mathcal T_U\setminus\{\emptyset\}$ below the chain. Rinse and repeat for these copies (taking only open subsets of the top $U$ for the copy), $\omega_1$ times. At the limits, the same thing is happening: any descending chain still corresponds to a bunch of open sets, so if their intersection is nonempty, add appropriate copies.

I think it's clear that if the players agree to play within the top $\mathcal T_X\setminus\{\emptyset\}$, the result is the same as the topological game. So we should argue that each player can enforce equivalent play to show strategies in the topological game translate here.

To see this, note that given however many moves of the opponent they're allowed to see, a player with a winning topological strategy/tactic can obtain an open set $U$ and choose the top element of a copy of $\mathcal T_U\setminus\{\emptyset\}$. (If they couldn't, then all infinite descending chains have empty intersection and the game is already set for P1.) The opponent then has to choose a subset of $U$ for the rest of the game.

So the winning player has ensured the game descends down our poset copies countably-far, and both players are essentially picking open subsets of each other. Let $\alpha<\omega_1$ be the least limit ordinal below this descent. If the open subsets have empty intersection, we didn't insert anything below that chain; otherwise, we did. This preserves the winning strategy/tactic for whichever player had one.

It remains to be shown that winning strategies in the modified game produce winning strategies in the topological game...

As JDH observed, there are topological spaces $X$ for which Player II has a winning strategy (and in fact a winning 2-tactic), but no winning tactic in the topological Banach-Mazur game. So we proceed by attempting to find a poset that mimicks the topological game.

Naively one might let choose $\mathcal T_X\setminus\{\emptyset\}$. But as observed by Will Brian, this isn't quite right: in the topological game, P2 doesn't win by finding an open set below all chosen sets; they win if these sets have non-empty (not necessarily open) intersection.

Another attempt might be to "add in" a minimal point at the bottom of a chain of open sets with non-empty but empty-interior intersection to mimic this win condition for P2. But then P2 would just pick this new valid move immediately and end the game.

So instead, for each such chain in $\mathcal T_X\setminus\{\emptyset\}$, take each open set $U$ that contains the non-empty intersection, and insert a copy of $\mathcal T_U\setminus\{\emptyset\}$ below the chain. Rinse and repeat for these copies (taking only open subsets of the top $U$ for the copy), $\omega_1$ times. At the limits, the same thing is happening: any descending chain still corresponds to a bunch of open sets, so if their intersection is nonempty, add appropriate copies.

I think it's clear that if the players agree to play within the top $\mathcal T_X\setminus\{\emptyset\}$, the result is the same as the topological game. So we should argue that each player can enforce equivalent play to show strategies in the topological game translate here.

To see this, note that given however many moves of the opponent they're allowed to see, a player with a winning topological strategy/tactic can obtain an open set $U$ and choose the top element of a copy of $\mathcal T_U\setminus\{\emptyset\}$. (If they couldn't, then all infinite descending chains have empty intersection and the game is already set for P1.) The opponent then has to choose a subset of $U$ for the rest of the game.

So the winning player has ensured the game descends down our poset copies countably-far, and both players are essentially picking open subsets of each other. Let $\alpha<\omega_1$ be the least limit ordinal below this descent. If the open subsets have empty intersection, we didn't insert anything below that chain; otherwise, we did. This preserves the winning strategy/tactic for whichever player had one.

It remains to be shown that winning strategies in the poset game produce winning strategies in the topological game. Given a winninng strategy for either player in the poset game, note that it must handle the situation where the opponent always chooses the top of a lower copy of $\mathcal T_U$ for each move. This enforces the winning strategy to essentially choose subsets of the opponent at each step. Then this can be translated down to the topological game. If the poset game resulted in no lower bound, it's because the intersection of open sets was empty; otherwise it's because the intersection of open sets was non-empty.