The problem is from Infinite prisoners with hats -- is choice really needed?
A countably infinite number of prisoners, each with an unknown and randomly assigned red or blue hat line up single file line. Each prisoner faces away from the beginning of the line, and each prisoner can see all the hats in front of him, and none of the hats behind. Starting from the beginning of the line, each prisoner must correctly identify the color of his hat or he is killed on the spot. The prisoners have a chance to meet and confer beforehand, but once in line, no prisoner can hear what the other prisoners say. The question is, is there a way to ensure that only finitely many prisoners are killed?
There's no strategy that guarantees $<N$ prisoners are killed, because we can imagine an outsider coming in before the game starts, and asking prisoner $N$ what they would guess, then changing their hat to the opposite color. Then do the same for prisoner $N-1$, and so on. This way the outsider guarantees the first $N$ prisoners dies.
This trick works because changing hats of prisoners $1$ to $N-1$ does not affect $N$. But now change the question so that every prisoner can see every other prisoner. The original strategy that guarantees only finitely many prisoners die still works, but now, is it possible to guarantee $< N$ prisoners would die, for some finite number $N$? (The prisoners still can't hear what others say.)
This is not possible.
Imagine again our outsider arbitrarily choosing $2N$ prisoners. The outsider can ask these prisoners what their guesses will be for each of the $2^{2N}$ ways the hats can be assigned to those prisoners (leaving the hats of the unchosen prisoners unchanged).
Each chosen prisoner will die in half of the assignments (choosing wrong in one of a pair in which the other hats are fixed), so there are $2N\cdot2^{2N-1}$ total deaths among all hat assignments. By the pigeonhole principle, then, there must be at least one assignment with at least $\lceil2N\cdot2^{2N-1}/2^{2N}\rceil = N$ deaths.
(Equivalently, the expected number of deaths is $N$, so there must be at least one assignment with at least $N$ deaths.)