Let n = 0.999... Then multiply both sides of the equality by 10 so that we have 10n = 9.999...Then subtract n from both sides of the resulting equality to get 9n = 9.000...Finally, divide both sides by 9 and voila we have n = 1 which is what we wanted to show.
My gut feeling is that that proof isn't quite correct, since you haven't used the notion of a limit anywhere. There's a fundamental fact about convergence of geometric series that you need to use.
I think your proof goes wrong since you haven't justified how arithemtic operations work with infinite decimals. AFAIK the only way to add non-terminating decimals is to convert them to fractions (or sequences of fractions as with pi, e, etc), add the fractions, and convert them back. So if you convert 0.999... and 9.999... to fractions, you've assumed the conclusion.
To play devil's advocate, I can try to rephrase your proof without infinite decimal arithmetic as follows.
Assume
n = 0.999... = 1 - epsilon, where epsilon is 'infinitesimal' (an ill-defined version of not-quite-zero). We'd like to show that epsilon is zero.
10n = 9.999 = 10 - 10epsilon
9n = 9.999 - (1 - epsilon) = 9 - 9epsilon
9n = 8.999 + epsilon = 9 - 9epsilon
The only way to get the epsilons to cancel is to assume epsilon = 0, which is to assume the conclusion.
