You spend a lot of time on what you call the "classic proof". However, what this is, is not a proof but a plausability argument.
The stadard approach start out
1. define the real numbers (e.g. Dedekind cuts)
2. let a = 0.999... be a real number. We do not need to give a full definition at this point a<=1 and a>(1-(1/10^n) for any natural number n is sufficient.
3. Note that since the real numbers form a field, 1-a must be a real number, call it d. Clearly d>=0, and d<(1/10^n) for any natural number n. The only real number with these properties is 0.
We conclude if 0.999... is a real number, then 0.999... = 1.
Note: under the standard "limit" definition of 0.999... = 1
You wish to call d a "dubious" number. However, if you use the usual definition of a real number then a dubious number is not a real number, nor is 0.999... = 1-d a real number. To make things work you have to abandon the usual real numbers and work with a system of "numbers" that is not a field.
