Other people have pointed out that induction never makes the jump from a finite number of 9s to an infinite number of 9s.

I feel the easiest "proof" is a proof by contradiction.

First hopefully we can agree that if we have two real numbers x and y that are not equal then we have a number z, such that x < z < y. The easiest example is z = (x + y)/2.

If 0.999...!= 1 then there must exist a number A, such that 0.999... < A < 1.

Now since 0 < A < 1 (as 0 < 0.999...) it should be easy to see that A's decimal expansion is of the form 0.abcdef... . Since, A != 0.999... one of the digits in the decimal expansion of A has to be something other than a 9.

For instance, we might have A = 0.99998999... .

However, this would mean A < 0.999... as all digits other than 9 are smaller than 9 [1], but this contradicts our initial assumption that 0.999... < A and since we're dealing with strict inequalities both can't be true at the same time! Thus no such A should exists and thus 0.999... = 1.

Now this isn't a rigorous proof, the thing that makes me most uncomfortable is the bit that I state A < 0.999..., but I'm uncomfortable because A might have multiple decimal expansions and I don't know how the algorithm in [1] interacts with that, however, if someone quibbles about that bit of the proof for that reason I feel they should have already accepted that 0.999... = 1 via another rigorous proof.

[1] If this is not clear think about how you would compare two decimal expansions to see if one is smaller than the other. You go through every digit until you find one that is different between the two numbers and then you compare those.