To some degree, this is true. 0.999... does represent the number closest to 1. That is, if we consider the interval (0,1) (which is all of the real numbers x such that 1 > x > 0 and x is not 0 or 1), 0.999... represents a the limit of a convergent sequence of numbers (0.9, 0.99, 0.999, ...) within this interval. Your thought is that since each of these approximations for 0.999... is less than *but not equal to* 1, 0.999... must also be less than (and not equal to) 1.

Sadly, this is not the case. For proof, let a and b be numbers. If b - a = 0, then by simple arithmetic, a = b. Hence by the contrapositive statement, we have that if a != b, then b - a != 0. Without loss of generality, assume that b is larger than a, so that b - a > 0. We can agree that 1 is not less than 0.999..., so in particular, we have that 1 != 0.999... if and only if 1 - 0.999... > 0.

Now, we know that for each approximation a_n of 0.999... (a_1 = 0.9, a_2 = 0.99, a_3 = 0.999, etc) 1 - a_n = .000...01 (where there are n-1 0s). And since 0.999... > a_n for each n, 1 - a_n > 1 - 0.999... for each n. In particular, this means that 10^{-n} > 1 - 0.999... for any natural number n.

Suppose for the sake of contradiction that 1 != 0.999.... Then 1 - 0.999... > 0, say 1 - 0.999... = e. Consider the reciprocal 1 / e. Because e > 0, 1 / e is some (possibly very large) number. Because 1 / e is a real number, we can find some n such that 10^n > 1 / e (This is the crux of the argument, which is sometimes called Archimedes's Principle. Most people agree that no matter how large a number you choose, I can choose another one that is bigger). But then e > 10^{-n}, so that 1 - 0.999... = e > 10^{-n} for some natural number n. This is a contradiction, since we found above that 10^{-n> > 1 - 0.999... for every natural number n. Because of this contradiction, one of our assumptions must be wrong, and our only non-obvious assumption was that 1 != 0.999.... Therefore 1 = 0.999...

The problem with your argument is that there is no number closest to 1 without being one. Just as I can always choose a number bigger than a given number, for any number less than 1 there is another number between that and 1.