Uncountability of $mathbb{R}$

I was trying to follow this proof of the uncountabilty of R and at first it seemed clear, but when I tried to explain it to myself I realized I didn’t really understand one of the steps.

The proof is by contradiction, using the nested intervals theorem:

Assume $mathbb{R}$ is countable. Then we can define a bijection from $mathbb{N}$ to $mathbb{R}$, in other words we can assign each number in $mathbb{R}$ a subscript in $mathbb{N}$ and get the infinite sequence $R = {x_1, x_2, x_3, … } $.

Let $I_1 subset mathbb{R}$ be a closed interval such that $x_1 notin I_1$. $I_1$ can also be written as $[a_1, b_1]$, with $x_1 < a_1 < b_1$.

So far so good. Now comes the step where I ran into difficulties.

Let $I_2 subset I_1$ be a closed interval such that $x_2 in I_1$ and $x_2 notin I_2$. $I_2$ has bounds $a_2, b_2$. This means we now have the following inequality $x_1 < a_1 leq x_2 < a_2 < b_2 < b_1$.

But it seems to me that for this to make sense $a_1$ must be equal to $x_2$, since we have assumed that $mathbb{R}$ is countable and we know that $mathbb{R}$ is increasing. Otherwise we’ve just produced a new number in $mathbb{R}$ between $x_1$ and $x_2$ that the count sort of skipped.

Now obviously the point of the proof is to show precisely that $mathbb{R}$ is uncountable and that assigning natural subscripts to the real numbers won’t get you anywhere, but it feels like at that moment in the proof we’d be asssuming the conclusion.

The proof then goes on to define $I_{n+1}$ as a closed interval $subset I_n$ such that $x_{n+1} notin I_{n+1}$.

Once we have these intervals constructed such that $I_1 supset I_2 supset I_3 supset$ … we can apply the nested intervals theorem, which tells us that the intersection of these nested sets is non-empty, and get:
$$exists x in mathbb{R};such;that; x in ( cap I_n forall n in mathbb{N} ) $$

Since we have assumed $mathbb{R}$ is countable, we know that $x = x_m, m in mathbb{N}$. So if the intersection of the nested sets is non-empty it must contain a number of the form $x_m$.

But we have constructed our sets such that for any number $x_m$ there is a nested set $I_m$ that doesn’t contain it. So the intersection of the nested sets cannot contain any number of the form $x_m$. Which is the contradiction we were after.

It seems to me like the rest of the proof holds even if we require that $a_1 = x_2$ (and by extension $a_n = x_{n+1}$). What am I missing? Does the proof still hold even if $a_1$ is allowed to be $leq x_2$? If so, why?