I know another good 'proof':
Proposition:
In any finite set of women, if one of them has blue eyes, they all do.
Proof:
We will prove this by induction.
IE, we prove that the proposition holds for 1, and we prove that if the proposition holds for a number n, it will also hold for n+1.
In this fashion, you know it works for 1, and therefore for 2 and for 3 etc etc.
-proof for 1 woman:
fairly trivial: in a set of one woman, if she has blue eyes, every woman in the set does
-proof for n+1
assume the proposition holds for n and consider a group of n+1 woman with at least one woman with blue eyes
line them up with a woman with blue eyes on the left:
B,x,x,x,x,x,.....,x
now the first n women will form a set of n women and one of them has blue eyes. Therefore, by our assumption all of them have blue eyes:
B,B,B,B,......,B,x
now look at the last n women. These women also form a set of women with at least one blue-eyes (in fact, the first n-1 women all have blue eyes) therefore, by our assumption, the last woman also has blue eyes and the entire set does
QED
(note: please don't try arguing against the method of proving by induction itself. It is a perfectly valid form of proof. Any flaws in this proof are in the specifics of this particular application of induction)