Proof that the reals are uncountable by Königsberger I found this proof in Chapter 2 (Satz 7) of "Analysis 1" by Königsberger (hope it is not a duplicate)
It begins by defining an "Intervallschachtelung" (I don't know any good translation), which is a sequence $I_1,I_2,...$ of closed intervals (in the real line) such that:


*

*$I_{n+1}\subset I_n$ for every $n\in\Bbb N$

*For every $\epsilon\gt0$ there is one intervall $I_n$ with length $\vert I_n\vert\lt\epsilon$


The book gives the next as a principle: For every Intervallschachtelung in $\Bbb R$ there is a real number that belongs to every interval (Intervallschachtelungprinzip)
Now the proof: Let $\Bbb R=\{x_1,x_2,...\}$ be the countable set of every real number. Now we construct an Intervallschachtelung $(I_n)$ such that $$x_n\notin I_n$$ beginning by setting $I_1=[x_1+1,x_1+2]$ and getting $I_{n+1}$ from $I_n$ by recursion: divide $I_n$ in three equal parts and choose $I_{n+1}$ to be the one that doesn´t contain $x_{n+1}$. $(I_n)$ is an Intervallschachtelung and therefore there is a number $s\in I_n$ for every $n\in\Bbb N$ but if $s$ had been some $x_k$ in $\Bbb R$, then $x_k=s\in I_k$ which is a contradiction and thus $\Bbb R$ is uncountable
Now my question: isn't this the same as Cantor's diagonal argument? You are "making" a number different from the ones that you have listed, if so, why is it necessary the some unproven principle? (I haven´t found any proof anywhere)
Why is it important to divide each interval into three parts? I'm asking because in a previous Staz, two (equal) parts were enough. I think it only serves to avoid having $x_{n+1}$ right between the two parts but could two non-equal parts also do the job (given that you have a procedure of avoiding that)?
Thanks
 A: A a high level, this is the same as the diagonal argument: both have the general form:

Assume that $f$ is any function $\mathbb N\to\mathbb R$. Then here is a procedure to find a real number that is not in the range of $f$. (bla bla bla bla). Therefore $f$ is not surjective.
Since $f$ was arbitrary, we have concluded that there is no surjective function $\mathbb N\to\mathbb R$. In particular there is no bijection.

They differ in how the procedure for finding a number that is not hit works.
(But notice that if we change the construction you cite to divide into ten subintervals instead of three, what actually happens is very close to the diagonal argument).
The nested-intervals method was the first uncountability proof Cantor published in 1874. The decimal-based diagonal argument came later, in 1891, and is generally thought of as more instructive (especially in the modern age where elementary-school mathematics puts a lot more weight on decimals than on fractions).

Dividing the interval into exactly three equal parts is not particularly important. It's just the easiest way to make sure that there is one of the subintervals that doesn't contain $x_n$.
If we divided the interval $[a,b]$ in to only two parts $[a,\frac{a+b}2]$ and $[\frac{a+b}2,b]$, we would be stuck if $x_n$ happened to be $\frac{a+b}2$, because that is in both intervals. As you say, this is easy enough to remedy if you hit the problem: just shrink one of the intervals slightly. You can view the three-intervals division as simply the remedy that the author thought would be simplest to describe.
Note that it is never necessary to select the middle interval, so we could also describe the same construction as "for $I_n = [a,b]$, consider the two intervals $[a,\frac{2a+b}3]$ and $[\frac{a+2b}3, b]$ ..." -- but this is again (at least arguably) slightly more involved to write and understand than just saying "divide $I_n$ into three equal parts".

As MooS says in comments, the Intervallschachtelungsprinzip is not unproven, it is equivalent to the fact that the real numbers are complete, i.e. that any Cauchy series converges or that any bounded set has a supremum.
In particular we can find a number is the intersection as the limit of the sequence of left endpoints. This sequence is upwards bounded (by the right endpoint of the first interval), and non-decreasing; therefore it has a limit. And it is a simple exercise to see that this limit must be in each of the intervals.
It may seem that the fact that such a sequence has a limit is a "new unproven assumption" that is needed only for this variant of the proof. But in fact the very same assumption is used in the diagonal proof -- hidden inside the taken-for-granted fact that  every infinite sequence of decimals defines a real number.
