the pumping lemma game

Using the pumping lemma to prove non-regularity is best pictured as a two-player game, because the lemma is a chain of 'for all ... there exists ... for all' quantifiers, and proving its contrapositive means winning that game as the challenger. You are trying to show NO pumping length can possibly work; an imaginary adversary defends the claim that the language is regular. If you have a forced win for every choice the adversary makes, the language is not regular.

The moves, in order: (1) The adversary picks a pumping length p — you must beat ANY p, so treat p as an arbitrary unknown. (2) YOU pick a single string w in the language with |w| ≥ p, chosen cleverly so its structure is rigid (for a^n b^n the killer choice is w = a^p b^p). (3) The adversary splits w = xyz obeying |y| ≥ 1 and |xy| ≤ p — but since |xy| ≤ p and w starts with p a's, y must consist only of a's, so the adversary has no good options. (4) YOU pick a pump count i that breaks membership: pumping i = 2 turns a^p b^p into a^{p+|y|} b^p, which has more a's than b's, so it is NOT in the language. Contradiction; the language is not regular.

The two crucial discipline points are these: you do NOT choose p (the adversary does, so your argument must survive every p), and you DO choose w and i (so pick the most constrained string you can, and the pump count that most obviously breaks it). Mixing these up — choosing your own p, or letting the adversary choose w — is the single most common way a pumping proof goes wrong.