Clarification for the proof of uncountability of real numbers

Question

The following is from the Understanding Analysis 2nd ed., Stephen Abbot, page 28

If we let $x_1 = f(1), x_2 = f(2)$ and so on, then our assumption that $f$ is onto means that we can write $\mathbb{R} = \{x_1, x_2, \dots\}$ and be confident that every real number appears somewhere on the list. We will now use the Nested Interval Property to produce a real number that is not there. Let $I_1$ be a closed interval that does not contain $x_1$. Next, let $I_2$ be a closed interval, contained in $I_1$ that does not contain $x_2$. The existence of such an $I_2$ is easy to verify.

This is a pedantic question, but in this scenario, where we have first assumed that $\mathbb{R}$ is countable, how would we verify the existence of $I_2$? Just bluntly exclude $x_2$ from $I_1$? Given that most if not all things in the book so far have been proven, why can we just state this? Is it due to the countability assumption on $\mathbb{R}$?

Answer 1

The interval $I_1$ is of the form $[a,b]$ for some real numbers $a$ and $b$ such that $a<b$. If $x_2\notin[a,b]$, you can just take $I_2=I_1$. Now, suppose that $x_2\in[a,b]$. If $x_2\leqslant\frac{a+b}2$, you can take $I_2=\left[\frac{a+2b}3,b\right]$. And, if $x_2\geqslant\frac{a+b}2$, you can take $I_2=\left[a,\frac{2a+b}3\right]$.