Is Abbott's Proof of the Uncountabilty of Real Numbers too strong? In Stephen Abbott's "Understanding Analysis" book, he proves that the set of Real numbers is an uncountable set in theorem 1.4.11. My issue with the proof is that it seems too strong, as in it seems like you can prove that the set of Integers or Rationals are uncountable with the exact same proof, just replace Reals with Integers or Rationals. Where does he invoke properties of the Reals that the Integers and Rationals don't have? I know I have to be missing something...
Here's the theorem and proof, copied straight out of Abbott:
Theorem 1.4.11. The set R of Real Numbers is uncountable.
Proof.  Assume that there does exist a 1–1, onto function f : N → R. Again, what this suggests is that it is possible to enumerate the elements of R. If we let x1 = f(1), x2 = f(2), and so on, then our assumption that f is onto means that we can write
(1) R ={$x_1$, $x_2$, $x_2$, $x_2$,...}
and be conﬁdent that every real number appears somewhere on the list. We will now use the Nested Interval Property (Theorem 1.4.1) 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 $I_1$, which does not contain $x_2$. The existence of such an $I_2$ is easy to verify. Certainly $I_1$ contains two smaller disjoint closed intervals, and $x_2$ can only be in one of these. In general, given an interval $I_n$, construct $I_{n+1}$ to satisfy
(i) $I_{n+1}$ ⊆ $I_n$ and
(ii) $x_{n+}$ $\mathrel{{\epsilon}\llap{/}}$ $I_{n+1}$.
We now consider the intersection $\cap_{n=1}^{\infty}I_n$. If $x_{n_0}$ is some real number from the list in (1), then we have $x_{n_0} \mathrel{{\epsilon}\llap{/}} I_{n_0}$, and it follows that
$x_{n_0} \mathrel{{\epsilon}\llap{/}} \cap_{n=1}^{\infty}I_n$.
Now, we are assuming that the list in (1) contains every real number, and this leads to the conclusion that $\cap_{n=1}^{\infty}I_n =∅$. However, the Nested Interval Property (NIP) asserts that $\cap_{n=1}^{\infty}I_n \mathrel{{=}\llap{/}}∅$. By NIP, there is at least one $x ~\epsilon \cap_{n=1}^{\infty}I_n$ that, consequently, cannot be on the list in (1). This contradiction means that such an enumeration of R is impossible, and we conclude that R is an uncountable set.
 A: Abbot uses the nested interval property   that the countable intersection of nested intervals is non empty. 
BUT the intersection need not have any integers or rational numbers.  It merely needs at least one (and it could be just one) real number.
Rough outline:

Let $\{x_1,x_2...\}$ be a countable list of all real numbers. 
  Construct $I_n$ being closed intervals so that $x_i \not \in I_i$ and
  $I_{i+1} \subset I_i$.
By the 1.4.1 The Nested Interval property $\cap_{i=1}^{\infty} I_i \in
 \emptyset$ So there exists a $w \in \cap_{i=1}^{\infty}I_n$.  But $w
 \in \mathbb R$ so $w \in \{x_1, x_2 ....\}$. So $w = x_k$ for some
  $k$.  And $w=x_k \in \cap_{i=1}^{\infty}I_n \subset I_k$.  But $x_k \in I_k$ which is a contradiction.

End of rough outline.
So if $\{x_1, x_2....\}\ne \mathbb R$ (or more precisely if $\mathbb R \not \subset \{x_1,x_2....\}$), we can only conclude that $w \in \mathbb R$ and we  can not conclude $w \in \{x_1, x_2....\}$.  
And thus we have no contradiction.
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Addendum:  This post is assuming that or space is still the reals and that it is only the list that is all the rationals or integers.
If we were to do "Rational Analysis" the proof fails because the rationals do not have the Nested Interval Property.  If you have nested series of $[a_n, b_n] = \{q \in \mathbb Q|a_n \le q \le b_n\}$ and $I_{n+1} \subset I_n$ then it is not true that $\cap I_n \ne \emptyset$.  
The rationals do not have the Axiom of Completeness and thus the set $\{x_i\}$ so that $a_i < x_i < a_{i+1}$ although increasing and bounded above need not have a least upper bound and $\cap I_n$ may be empty.
Example:  Let $\{[a_i, b_i]\}$ so that $a_i^2 < 2- \frac 1n < a_{i+1}^2 < 2 < b_{i+1}^2 < 2 + \frac 1n < b_{i}^2$ and $a_, b_i$ are positive.  Then $\cap [a_i,b_i] = \emptyset$.
The integers do have the Nested Interval Propety but If you have a sequence of $I_{i+1} \subset I_{i}$ and $I_i \ne \emptyset$ then the is an $N$ so that $I_i = I_N$ for all $i > N$.  As such the sequence of nested intervals as described in the proof of 1.4.11 can not be created.   Ultimately you will not be able to create $I_{i+1} \subset I_i$ with $x_{i+1} \not \in I_{i+1}$.
The "The existence of such an I2 is easy to verify" is not possible in the ingeters.  In the reals (or the rationals)one can verify that in $(a_i, b_i)$ there are $a_{i+1}, b_{i+1}$ so that $a_i < a_{i+1}<  b_{i+1}$ and if $x_{i+1} \in I_i$ we can avoid it by selecting proper $a_{i+1} , b_{i+1}$.  
In the integers that is simply not the case.  Give $a_{i} < b_i$ there only finitely many (possible none) $a_{i+1}, b_{i+1}$ so that $a_{i} < a_{i+1} < b_{i+1} < b_i$ so this can not be extended.  Eventually there will have to be an $x_n \in I_n$. 
