What's wrong with this “proof” that $\mathbb{R}$ is enumerable?

Question

The fake proof:

• We know that $\mathbb{R}$ is uncountable, hence we cannot enumerate over it.
• But what we do know is that $\mathbb{Q}$, the set of rationals, is countable, and even denumerable.
• We also know that we can construct $\mathbb{R}$ through what are called Dedekind cuts.
• We choose to let the partition itself denote a new number and go forth to define mathematical operations on it as to be compatible with the rest of the numbers (mainly $\mathbb{Q}$ and our new number $x$)

Sidenote: I think so far this is standard, and contains nothing false. The actual argument starts below this line.

• Let us denote the set containing $x$ as $S_1 := \mathbb{Q}\cup\{x\}$. For convenience, the superscript of $S_1$ is how many new such numbers we have added through the cuts.

• Since $\mathbb{Q}$ is countable, we can enumerate over every single rational $q\in\mathbb{Q}$ to produce an $r\in\mathbb{R}$. Do this process $n$ times and you end up with $S_n = \mathbb{Q}\cup{x_1}\cup{x_2}\cup\dots\cup{x_n}$.

• But $S_n$ is also enumerable since it has a finite more elements than $\mathbb{Q}$.

• Hence - After enumerating over the entirety of $\mathbb{Q}$ - Start enumerating over the entirety of $S_{|\mathbb{N}|}\setminus\mathbb{Q}$

• Now we will end up with even newer numbers to put in our set, which we will now call $S_{n = |\mathbb{N}|,k}$ where $n$ represents the enumeration over $\mathbb{Q}$ and $k$ represents the enumeration over $S_{|\mathbb{N}|}\setminus\mathbb{Q}$. Do this ad infinitum and you will eventually describe $\mathbb{R}$.

I know I went wrong somewhere, I just don't know where.

Answer 1

"Do this ad infinitum and you will eventually describe $\mathbb{R}$."

The "ad infinitum" takes uncountably many steps to complete.