How do we know that Cantor's diagonalization isn't creating a different decimal of the same number?

Solution 1:

This is in fact a potential problem if the proof is carelessly stated, but it’s easily avoided: if the $n$-th decimal digit of the $n$-th number on the list is $7$, we replace it by $6$, and if it’s not $7$, we replace it by $7$. The only numbers in $(0,1)$ with two decimal representations are those with one representation ending in an infinite string of nines and the other in an infinite string of zeroes, and this version of the argument clearly doesn’t produce a number of either of those forms.

Solution 2:

As Henning Makholm stated in his comment, you're supposed to select digits that are different from the diagonal you look at, not the diagonal digits themselves. Make sure you also select digits different from $9$ and $0$, and the problem is avoided