Edit: As the comments mention, I misunderstood how to use the diagonalization method. However, the issue I'm trying to understand is a potential problem with diagonalization and it is addressed in the answers so I will not delete the question.
Cantor's diagonalization is a way of creating a unique number given a countable list of all reals.
I can see how Cantor's method creates a unique decimal string but I'm unsure if this decimal string corresponds to a unique number. Essentially this is because $1 = 0.\overline{999}$.
Consider the list which contains all real numbers between $0$ and $1$:
$0.5000 \mathord\ldots \\
0.4586 \mathord\ldots \\
0.3912 \mathord\ldots \\
0.3195 \mathord\ldots \\
0.7719 \mathord\ldots\\
\vdots$
The start of this list produces a new number which to four decimal places is:
$0.4999 \mathord\ldots$
But $0.5$ was the first number and $0.4\overline{999} = 0.5$ so this hasn't produced a unique number.
Of course my list is very contrived, I admit that it's hard to imagine a list of the reals where numbers would align nicely to give a problem like this (since some numbers have no nines). However, I can't see a good reason why such an enumeration of numbers would be impossible.
Best Answer
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.