Let us have the following game. The player $F$ claims that the usual pigeonhole principle is not true and, moreover, they have a counterexample, i.e. a function which maps $n + 1$ pigeons injectively into $n$ holes for some finite $n$. The player $T$ does not agree with such a claim and wants to prove $F$ wrong. To achieve this, they ask $F$ questions of the following form: given a pigeon $a$, where should $a$ be mapped according to $F$. The player $T$ wins as soon as the $F$'s answer contradicts some information $T$ already has, i.e. $F$ claims that the pigeon $a$ should be mapped to the hole $b$, while $T$ already knows that $a$ is mapped to the hole $b' \neq b$. The problem is that $T$ has really small memory, in particular, they can store information about $2$ pigeons at the same time, and so before asking any new question, they must erase the oldest piece of information (and so immediately prior to asking such a question $T$ effectively has information about $1$ pigeon only).
It is not hard to show that that in this game $F$ can play in a way that $T$ will never catch them off guard and so $T$ can't possibly win. But there is one more rule. At some point $T$ can ask the last question, after which they will take the final piece of information and compare it with all the data they have gathered through the whole game (the data which has been erased). Note that they can compare the answers gathered through the whole game with the last answer only and not with each other. If they can find some contradiction with the $F$'s claim, they win.
Assume $T$ has some fixed strategy (i.e. each new question depends only on information which is in $T$'s memory and on the question's index) and assume the index of the final question is fixed as a part of the strategy. I would like to find an (elementary) proof that $F$ can still play in such a way that $T$ will not win in the end.
If F knows which pigeon (call it X) will be asked about in the final question, they can fool T into thinking that they can fit any number of pigeons into just 3 holes.