If you accept that
1. all numbers can be represented by infinite decimal expansions in the form of a function f: N->{0,1,2,3,4,5,6,7,8,9, .}, (for example, 0.99999... is the function f(0)=0, f(1)=. , f(2)=9, f(3)=9, f(n)=9 for all n>3 . 1 can be represented by g(0)=1, g(1)=., g(2)=g(3)=... = g(n) = 0). We assume that every representation determines a unique number but not necessarily that every number has a unique decimal representation.
2. you can do arithmetic on numbers in expected ways,
then you are forced to have
0.9999....=1
Proof:
By comparing each digit, we clearly see that 1 is not less than 0.999....
Now subtract the two numbers (which you can do by assumption 2)
1-0.999... = x >= 0
As every number can be represented by an infinite decimal expansion (assumption 1), there is a sequence of digits for x.
If there are any nonzero digits of x after the dot, there is a first nonzero digit and it has a well-defined natural number position N. But clearly that cannot be, since then it would mean that x is bigger than some fraction 1/10^N. Thus every digit of x must be zero and so x itself is zero.
So 1-0.999... = 0
and hence
0.999... = 1
QED