Using nested intervals to prove that $\mathbb{R}$ is not countable I am having trouble understanding this proof from the book "Understanding Analysis" by Stephen Abbot. The problematic part is on page 29 of the book, which is available on google books.
Assuming that $\mathbb{R}$ is countable, it can be represented as a sequence of numbers:
$\mathbb{R}_s = \{ x_1, x_2, x_3, ... \}$
The idea of the proof is to use nested interval theorem to show that there exists $x \in \mathbb{R}$ that is not an element of the sequence $\mathbb{R}_s$. This would mean that no such sequence of real number exists.
The proof goes on to construct finite and nested intervals $I$ such that: 
(1) $x_{n} \notin I_n$ and 
(2) $I_{n+1} \subseteq I_n$. 
It is assumed that $\mathbb{R}_s$ contains all real numbers. Since $x_{n} \ne I_n$, it is clear that for the first number we choose
(3) $x_{n0} \notin\cap_{n=1}^\infty I_n$
However, the proof claims that since $\mathbb{R}_s$ contains all real numbers, this leads to $\cap_{n=1}^\infty I_n = \emptyset$
This contradicts with the nested interval theorem that states the opposite: $\cap_{n=1}^\infty I_n \ne \emptyset$
Therefore no such sequence $\mathbb{R}_s$ can be constructed, which means $\mathbb{R}$ is not countable. 
My questions
Q1: If all the intervals are finite, how can we even use $\cap_{n=1}^\infty I_n$? 
Example: Choose $x_3$ for $x_{n0}$ and $\{x_4, x_5, .... x_{n}\}, n \in \mathbb{N}$ for $I_1$ - following (1) and (2). How can I construct infinitely many nested intervals from a finite $I_1$ following (1) and (2)? There are maximally $n-1$ such nested sub-intervals, if $I_{n+1}$ is constructed from $I_n$ such that the first or the last element of $I_n$ is chosen as $x_{n+1}$.
Q2: This part:

(3) $x_{n0} \notin\cap_{n=1}^\infty I_n$
However, the proof claims that since $\mathbb{R}_s$ contains all real
  numbers, this leads to $\cap_{n=1}^\infty I_n = \emptyset$

(3) is perfectly clear, but how does that lead to $\cap_{n=1}^\infty I_n = \emptyset$? I mean, sure, (3) is valid for any $x_{n0} \in \mathbb{R}_s$, but that doesn't break the nested interval theorem. The nested interval theorem states basically that $I_n \ne \emptyset$. 
Is this the point? Rules (1) and (2) will enforce a constructuion of such $I_{n}$ that is an empty set, causing the intersection of all intervals to be empty?
 A: First of all, rather than $x_n\ne I_n,$ $(1)$ should read $x_n\notin I_n.$ What would it even mean for a real number to be equal to an interval of real numbers? This may be mistranscribed by you, or it may be a mistake in your text.

Second of all, there is the statement "there are maximally $n-1$ nested subintervals, if $I_{n+1}$ is constructed from $I_n$ such that the first or the last element of $I_n$ is chosen as $x_{n+1}.$" I suspect that you're misunderstanding what is meant by "finite" in the sense of intervals. It does not mean that the interval has only finitely-many elements. Rather, it means that the interval is bounded (or, equivalently, of finite length). In particular, each interval in the construction  has the form $[c,d]:=\{x\in\Bbb R:c\le x\text{ and }x\le d\}$ for some $c,d\in\Bbb R$ with $c<d$. I'm not sure how your proof constructs these intervals, but here is one way we can proceed:

First of all, let $I_1=[x_1+1,x_1+2].$ Note that this is a finite, non-empty, closed interval, and that $x_1\notin I_1.$ Suppose we have a finite, non-empty, closed interval $I_n.$ Then there are a few cases to consider.

*

*If $x_{n+1}\notin I_n,$ then we simply let $I_{n+1}=I_n$.


*If $x_{n+1}$ is the least element of $I_n$--meaning that $I_n=[x_{n+1},d]$ for some $d>x_{n+1}$--then we choose some $c$ with $x_{n+1}<c$ and $c<d$--for example, we could use $c=\frac12(x_{n+1}+d)$--and let $I_{n+1}=[c,d].$


*If $x_{n+1}$ is an element of $I_n,$ but is not the least element--meaning that $I_n=[c,d]$ for some $c<x_{n+1}$ and some $d\ge x_{n+1}$--then we choose some $d'$ with $c<d'$ and $d'<x_{n+1}$--for example, $d'=\frac12(c+x_{n+1})$--and let $I_{n+1}=[c,d'].$

You should be able to see/prove that, regardless of which case we have $I_{n+1}$ is a finite, non-empty, closed subinterval of $I_n$ such that $x_{n+1}\notin I_{n+1}.$ Note, however, that there were many ways we could have chosen the subinterval! For example, in the first case, we could have chosen any (non-empty, closed) subinterval of $I_n$. In the second case, we could have chosen $c$ in many ways--for example, given any $0<t<1,$ we could let $c=tx_{n+1}+(1-t)d.$ Similarly, we could choose $d'$ in many ways in the third case. (It turns out that there are uncountably-many ways to choose an appropriate sub-interval, but that isn't important, and is tantamount to what we're trying to prove!) Moreover, the above procedure recursively constructs all of the $I_n$ for us.

Finally, there is the statement "(3) is valid for any $x_{n_0}\in\Bbb R_s,$ but that doesn't break the nested interval theorem. The nested interval theorem states basically that $I_n\ne\emptyset.$" But this is not so! Rather, the NIT states that if there is a sequence of intervals $I_n$ of real numbers such that (i) each $I_n\ne\emptyset,$ (ii) each $I_n$ is of the form $[c,d]$ for some real $c,d,$ and (iii) $I_{n+1}\subseteq I_n$ for all $n,$ then we have that $$\bigcap_{n=1}^\infty I_n:=\{x\in\Bbb R:x\in I_n\text{ for all }n\ge 1\}\ne\emptyset.\tag{$\star$}$$ Note that it is necessary that each $I_n$ is non-empty, but the conclusion is $(\star).$ Since our construction meets the criteria for NIT, we can conclude that $(\star)$ holds, and so there is some $x\in\bigcap_{n=1}^\infty I_n.$ By assumption, $x\in\Bbb R_s,$ so $x=x_{n_0}$ for some $n_0.$ But by (3), this is impossible: since $x_{n_0}\notin I_{n_0}$ by construction, then $x=x_{n_0}\notin\bigcap_{n=1}^\infty I_n$. Thus, we have a contradiction.
A: First of all I think finite in this case is supposed to mean bounded. You can't have finite intervals in $\mathbb{R}$ in terms of the number of points contained in the intervals with the exception of one point intervals.
As for the proof I think it's fairly straightforward. You take an arbitrary closed interval a the start. Let's say $[0,1]$.
At each step you define $I_{n+1}$ to either be $I_n$ if $x_n\not\in I_n$ or assuming $I_n=[a_n,b_n]$ you split the interval into thirds $I_{n,1}=[a_n,a_n+(b_n-a_n)/3]$,$I_{n,3}= [a_n+2(b_n-a_n)/3,b_n]$ and choose an $\alpha$ so that $x_n\not\in I_{n,\alpha}$ (you can always choose one of these since the two intervals are disjoint).
These intervals are then nested non-empty and closed thus $\bigcap I_n\neq\emptyset$, thus $\exists x\in\bigcap I_n$. But this $x$ is real and so was enumerated in our original sequence $\mathbb{R}_s$ and thus at some stage $n$ we had $x\not\in I_n$, thus $x\not\in \bigcap I_n$ and we have a contradiction.
A: For Q2:
We have shown that for all $n0$, $x_{n0}\notin \cap_{n\geq 1}I_n$. That is, for all $x\in\mathbb{R}_s$, $x\notin \cap_{n\geq 1}I_n$. Since we assume $\mathbb{R}_s=\mathbb{R}$ this implies $\cap_{n\geq 1}I_n=\emptyset$
For Q1:
Since we assume $\mathbb{R}$ is countable, we can consider a sequence $(x_n)_n$, with $x_1>1$ which includes all the real numbers. Let's construct the sets $I_n$ recursively like this: $I_1=[0,1]$ (then $x_1\notin I_1$), and given $I_n$, let 
\begin{equation*}
I_{n+1}=\begin{cases}
I_n &\text{if $x_{n+1}\notin I_n$}\\
[\min I_n,(\min I_n+x_n)/2]\subset I_n &\text{if $x_{n+1}\in (\min I_n,\max I_n]$}\\
[(\min I_n+\max I_n)/2,(\max I_n)]\subset I_n &\text{if $x_{n+1}= \min I_n$}
\end{cases} 
\end{equation*}
This sequence will be such that conditions (1) and (2) are verified.
