Question about counter-argument to Cantor's Diagonalization Argument.

For one of my homework assignments I was given the following complaints about his argument:

Every rational number has a decimal expansion so we could apply the Cantor Diagonalization Argument to show that the set of rational numbers between 0 and 1 is also uncountable.However, because we know that any subset of Q must be countable, there must be a flaw in Cantor’s Diagonalization Argument.

My idea is that the flaws in this argument are the following: Firstly, the set of numbers between 0 and 1 does not only consist of rationals - there are irrational numbers as well. Therefore, the argument that the set of numbers between 0 and 1 is rational and therefore countable is incorrect. However, I might be misunderstanding the problem because it is talking about the set of rationals between 0 and 1 specifically, so I'm just generally confused.

Any hints or suggestions would be greatly appreciated, thank you!

Edit: After reading the comments, I understand my initial thoughts above were incorrect. So, since this is looking at the rationals in the set (0,1), would I be correct in saying that if we used Cantor's method, there is no way of ensuring that the diagonal taken would also be a rational (since a rational can be written as a repeating decimal expansion). Since the diagonal might not ever repeat (because in theory, each added number in the diagonal can continue infinitely and never repeat). Would this argument be correct?

• You are misreading the question. The statement is not that "the set of numbers between $0$ and $1$ is rational and therefore countable," but rather that "the set of rational numbers between $0$ and $1$ is countable." – Théophile Apr 23 at 3:34
• The argument is: “take all rational numbers between $0$ and $1$. Create a list of them. Apply Cantor’s Diagonalization argument to this list, and thus exhibit a rational between $0$ and $1$ that is not in your original list. Thus, the collection of rational numbers between $0$ and $1$ is uncountable.” Your “firstly” is then nonsense; while correct in your assertion, that assertion is irrelevant because you are not asserting all numbers in $(0,1)$ are rationals. The real problem with the ‘argument’ proposed is elsewhere. – Arturo Magidin Apr 23 at 4:03
• Yes, your new reasoning is correct. Well done! (You could write this up as an answer below.) – Théophile Apr 23 at 12:14