This is only a partial answer. ZF + DC + 'every Banach-Mazur game is determined' is inconsistent.

Let $X$ be the set of all functions of the form $f: \alpha \rightarrow \{0,1\}$, with $\alpha$ an ordinal such that for any ordinals $\beta < \gamma$ with $\omega \cdot \gamma + \omega \leq \alpha$, the sets $(n \in \mathbb{N} : f(\omega \cdot \beta + n) = 1)$ and $(n \in \mathbb{N} : f(\omega \cdot \gamma + n) = 1)$ are distinct.

Let $\tau$ be the topology on $X$ generated by sets of the form $U_f = (g \in X : g \supseteq f)$, and let $\mathcal{X} = (X,\tau)$.

Claim: Player $1$ does not have a winning strategy for the Banach-Mazur game on $\mathcal{X}$.

Proof of claim: For any strategy $S$ for player $1$. Let $T$ be the tree of all initial segments of plays against $S$ such that each play by player $2$ is of the form $U_f$ for some $f \in X$ of length $\omega \cdot \alpha$ for some $\alpha$. If this tree fails to be pruned, then player $1$ has in some play of the game played a set $V$ such that for any $U_f \subseteq V$, $f$ enumerates every element of $2^{\mathbb{N}}$ extending $\sigma$ for some $\sigma \in 2^{<\omega}$. This would imply that $2^{\mathbb{N}}$ can be well-ordered, allowing us to construct a Bernstein set, contradicting our assumptions. Hence this must be a pruned tree of height $\omega$, so by dependent choice it has a path. Let $g$ be the union of all $f$ such that $U_f$ is on that path somewhere. By construction, $g \in X$, so we have that the strategy where player $2$ blindly plays the moves in this path wins against $S$.

So, it must be the case that player $2$ has a winning strategy $S$. For any $f \in X$ with length $\omega \cdot \alpha$ for some $\alpha$, let $T_f$ be the strategy for player $1$ that plays $U_f$ on player $1$'s zeroth move and on player $1$'s $n+1$st move, if player $2$ played $V$, then player $1$ plays $V \cap \bigcup_{\sigma \in 2^n} U_{f\frown \sigma\frown 0}$, if this is non-empty, and otherwise player $1$ plays $V \cap \bigcup_{\sigma \in 2^n} U_{f\frown \sigma\frown 1}$. (Note that since the union of these two is $V$, one of these must be non-empty.) Since $S$ is a winning strategy, for any $f\in X$ of length $\omega \cdot \alpha$ for some $\alpha$, the play of $T_f$ against $S$ must result in a nonempty intersection. By construction, for any $g$ and $h$ in that intersection, $g(\omega \cdot \alpha + n) = h(\omega \cdot \alpha +n)$ for all $n<\omega$, and the set $(n \in \mathbb{N} : h(\omega \cdot \alpha + n) = 1)$ must be distinct from $(n \in \mathbb{N} : f(\omega \cdot \beta + n) = 1)$ for any $\beta < \alpha$.

Therefore $S$ gives us a uniform procedure for picking a real not on a given well-ordered list of reals. By iterating this gives us a well-ordering of the reals. Therefore we can construct a Bernstein set from $S$ and we have that the Banach-Mazur game on that set is not determined, which contradicts our assumption. Therefore ZF + DC proves that there is an undetermined Banach-Mazur game.

Right now I don't see how to use a failure of DC to build an undetermined game.