I am reading Kechris' descriptive set theory text book, and there is this Theorem regarding infinite games:
Gale-Stewart: Let $T$ be a non-empty pruned tree on $A$. Let $X\subset[T]$ be closed or open. Then $G(T,X)$ is determined.
There is this exercise: show that AC is equivalent to Gale-Stewart Theorem.
Assuming the exercise is true, if we consider the statement:
$\blacksquare$ Every infinite game is determined.
Then the statement $\blacksquare$ implies Gale-Stewart Theorem, which implies AC, which implies the negation of $\blacksquare$.
Does this mean ZF$+\blacksquare$ is inconsistent? Can we conclude that we can find a non-determined infinite game without using AC? If so, do you have any example in your mind?
Thanks!
Your reasoning is correct: $\mathsf{ZF}$ proves $\lnot \blacksquare$.
Note that $\blacksquare$ is not $\mathsf{AD}$, the Axiom of Determinacy. $\mathsf{AD}$ states that every Gale-Stewart game on $\omega^\omega$ is determined, which is much weaker than $\blacksquare$.
It is possible to prove in $\mathsf{ZF}$ alone that there is a game on $\omega_1^\omega$ which is undetermined ($\omega_1$ is the first uncountable ordinal). A reference is Theorem 10.2 in Games with perfect information by Mycielski, Chapter 3 in Handbook of Game Theory with Economic Applications, 1992, vol. 1, pp 41-70 (link).
For a more explicit example of an undetermined game in $\mathsf{ZF}$, see the answer here by Joel David Hamkins.