TITLE: How do we know that Cantor's diagonalization isn't creating a different decimal of the same number?
QUESTION [90 upvotes]: 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.

REPLY [2 votes]: Cantor's Diagonal proof was not about numbers - in fact, it was specifically designed to prove the proposition "some infinite sets can't be counted" without using numbers as the example set. (It was his second proof of the proposition, and the first was criticized because it did.)
What Cantor actually used, were infinite length strings that combine only two different characters. He used "m" and "w." But modern High School and Middle School students have experience with infinite decimal expansions, and it is easier to teach new concepts to them in terms of what is already familiar. The point is, that there is no ambiguity in saying that two strings are different if they have different characters in at least one position.
Another point that is taught poorly, is that the proof doesn't have to be a proof-by-contradiciton. Many people have a justifiable objection to assuming statement A, and then proving that not(A) follows from it. All it really proves is that there is something wrong in the circle that leads from A to not(A), not that it has to be the assumption of A. Which is why so many people try to invalidate one of the steps - what you claim here one of the ways they try (and yes, it can be gotten around easily, as stated in other answers).
Cantor only assumes that you have a countably infinite set of some of these strings. Diagonalization proves that there is a string not in your set. Since "If A then B" is logically equivalent to "If not(B) then not(A)," this proves that if you have a set of all such strings, then it can't be counted.
Elegant, no?