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_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,k}$ where $ n$ represents the enumeration over $ \mathbb{Q}$ and $ k$ represents the enumeration over $ S_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.