Here is another partial answer.

Theorem. Suppose Player II has a winning strategy $\sigma$ in the Banach–Mazur game on a poset $P$. If there is a subset $D\subseteq P$ such that
(1) $\forall p\in P\ \exists d\in D\ d\le p$ and
(2) $\forall p\in P\ |\{d\in D:d\ge p\}|\le\aleph_0$,
then Player II has a winning tactic.

Proof. Suppose $p\in P$; I have to define $\tau(p)\le p$.

Let $S_p$ be the set of all partial $\sigma$-plays $u_1\ge v_1\ge\cdots\ge u_k\ge v_k\ge p$ ($k\in\omega$) (so that it is Player I's turn to muve, and anything $\le p$ is a legal next move) with all $u_i\in D$. Note that $S_p$ is countable; fix an enumeration $S_p=\{s_{p,n}:n\in\omega\}$. Construct an infinite sequence $$p\ge d_0\ge x_0\ge y_0\ge d_1\ge x_1\ge y_1\ge\cdots\ge d_n\ge x_n\ge y_n\ge\cdots$$ satisfying the conditions:
(i) $d_n\in D$;
(ii) $s_{p,n}$ followed by $d_n$ and $x_n$ is a partial $\sigma$-play;
(iii) $y_n=\sigma(x_0,x_1,\dots,x_n)$.

Since the sequence $x_0,y_0,x_1,y_1,\dots,x_n,y_n,\dots$ is a $\sigma$-play, it is bounded below; choose a lower bound and call it $\tau(p)$.

Any $\tau$-play is interlaced with a $\sigma$-play and therefore bounded below, i.e., $\tau$ is a winning tactic.

Here is another partial answer.

Theorem. Suppose Player II has a winning strategy $\sigma$ in the Banach–Mazur game on a poset $P$. If there is a subset $D\subseteq P$ such that
(1) $\forall p\in P\ \exists d\in D\ d\le p$ and
(2) $\forall p\in P\ |\{d\in D:d\ge p\}|\le\aleph_0$,
then Player II has a winning tactic.

Proof. Suppose $p\in P$; I have to define $\tau(p)\le p$.

Let $S_p$ be the set of all partial $\sigma$-plays $u_1\ge v_1\ge\cdots\ge u_k\ge v_k\ge p$ ($k\in\omega$) (so that it is Player I's turn to move, and anything $\le p$ is a legal next move) with all $u_i\in D$. Note that $S_p$ is countable; fix an enumeration $S_p=\{s_{p,n}:n\in\omega\}$. Construct an infinite sequence $$p\ge d_0\ge x_0\ge y_0\ge d_1\ge x_1\ge y_1\ge\cdots\ge d_n\ge x_n\ge y_n\ge\cdots$$ satisfying the conditions:
(i) $d_n\in D$;
(ii) $s_{p,n}$ followed by $d_n$ and $x_n$ is a partial $\sigma$-play;
(iii) $y_n=\sigma(x_0,x_1,\dots,x_n)$.

Since the sequence $x_0,y_0,x_1,y_1,\dots,x_n,y_n,\dots$ is a $\sigma$-play, it is bounded below; choose a lower bound and call it $\tau(p)$.

Now any $\tau$-play is interlaced with a $\sigma$-play and therefore bounded below, i.e., $\tau$ is a winning tactic.