Suppose a number of the form 0.000...1 is in fact nonzero

1. All such numbers are less than 1.
2. Real numbers have a least upper bound property. If a set of reals has an upper bound, then that set has a least upper bound.
3. Given (1) and (2) we conclude there is some number x, greater than 0 but less than 1, that is the least upper bound of all numbers of the form 0.000...1.
4. Let's call those numbers of the form 0.000...1 "irregular" numbers.
5. Conversely, any real number with a finite number of leading zeros in its decimal representation is a "regular" number. That includes even real numbers greater than or equal to 1.
6. The product of regular numbers is regular: the number of leading zeros of such a product is finite.
7. The product of a regular number and an irregular number is irregular: adding or subtracting a finite number of leading zeros doesn't change that the number of leading zeros in an irregular number is infinite.
8. Given (3), we ask if x--the least upper bound of all irregular numbers--is itself regular or irregular.
9. If x is regular, then 0.5x is also regular.
10. But 0.5x must also be less than some irregular number; otherwise it would be the actual least upper bound.
11. A positive regular number, with a finite number of leading zeros, can't be less than any positive irregular number. That's a contradiction.
12. If x is irregular, then 2x is also irregular. That also contradicts the concept that x is itself the least upper bound of all irregular numbers.

Hence "irregular" numbers as described in (4) are inconsistent with the least upper bound property of the reals.