It'd be like proving x^2 - 1 = (x+1)(x-1) for all x by starting with 1 and walking up the integers... Which is to say, not a bad way to prove it, but not necessary when you can logic it out otherwise.
Suppose there are only three people with blue eyes. A knows that B can see at least one blue-eyed person. And B knows that C can see at least one blue-eyed person. But A doesn't know that B knows that. A reasons: if I have brown eyes, B can only see one blue-eyed person, ie C. And if B can only see one blue-eyed person, he will reason "if I have brown eyes, C cannot see anybody with blue eyes". Thus, if I have brown eyes, B does not know whether C can see anybody with blue eyes.
Or when there are four blue-eyed people, A's reasoning will be: if I have brown eyes, then B will only be able to see two blue-eyed people. He will then reason according to the above case. So if I have brown eyes, B doesn't know that C knows that D can see at least one blue-eyed person.
By saying "I see a blue-eyed person", the foreigner removes this uncertainty.
"B does not know whether C can see anybody with blue eyes." - that's not true. C can see A.
And for the four person case, the same is true. If there is more than one blue eyed person, everyone knows there's at least one blue-eyed, person, so no new information via the foreigner.
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edit for the three person all blue eyes walk through, via the eyes of person A. Persons B and C have the same exact thoughts.:
Day 0) Everyone gains consciousness together. I look around and see two people with blue eyes. Either I have blue eyes or not blue eyes, and I'll know soon enough which based on how confused the others are.
Day 1) So everyone's still here. Great. They're confused too. Well, that's to be expected. B and C saw each others' blue eyes, but couldn't know whether or not their own matched, even if I were not blue eyed. Maybe they saw my blue eyes and needed even more time? Doesn't matter, tomorrow's D-day.
Day 2) Oh shoot. Everyone's still here. If I were not blue eyed, these guys would've noticed yesterday that something was up and figured out that they individually had blue eyes. I guess that means I confused them an extra day, so I have blue eyes, too.
Day 3) [silence]
Also note: the game not only progresses without the outsider, but it relies on the fact more than one person shares an eye color.
Yes. But what you quoted was part of A's reasoning. A doesn't know that he has blue eyes. If he doesn't, then C can only see B with blue eyes. And if C can only see B, then B doesn't know if C can see anyone. So A doesn't know whether B knows C can see a blue-eyed person (implying A has blue eyes) or not (implying he doesn't).
It's not about what people know directly. It's about what people know about other people's knowledge. In the three-person case, I know that there is at least one blue person. I know that everyone knows it. But I don't know that everyone knows that everyone knows it.
If K(0) is "I know there is at least one blue-eyed person", and K(n) is "I know that everyone knows K(n-1)", then (assuming I have blue eyes), K(0) is false only if I am the only blue-eyed person. K(1) is false if there are one or two of us. In general, K(n) is false if there are n+1 or fewer people with blue eyes. The foreigner makes K(n) true for all n.
The original puzzle says that there are an unknown number of eye colors (which happens to be just blue or brown, but the islanders don't know it), and the only reason blue-eyed people are killing themselves is because of the foreigner's statement; note that none of the brown-eyed people kill themselves in the inductive solution, because one of them could be (e.g.) green-eyed but not know whether or not they are the only one. The fate of the brown- and blue-eyed people would be reversed if the foreigner said "I see a brown-eyed person" instead.
As for the more subtle errors, let's assume there are only blue or brown eyes and the islanders know this. Now suppose A is brown and B and C are blue. Then B and C each see one blue and one brown. I think you'd agree that without a foreigner coming to the island and telling them that blue exists, neither can tell which their own color is. But A sees two blue -- exactly the same as in your example of three blue. So if A is using your reasoning: Day 0: no one dies. Day 1: no one dies. Day 2: no one dies. Day 3: B and C still have no idea what their color is, but A, seeing the same eyes as the A in your example and no one dead yet, commits suicide believing he has blue eyes, even though he does not. Clearly A's logic is faulty.