TITLE: $\mathbb{R}$ is uncountable, Abbott's proof
QUESTION [6 upvotes]: There is this proof in Abbott's book (Understanding Analysis) on page 25, that I am failing to understand.
Theorem is that $\mathbb{R}$ is uncountable. And here is how the author proceeds to prove it (I know there is an easier proof by Cantor):
To show that a set is uncountable, we need to show that there is no 1-1,onto function of the form: $f:\mathbb{N} \rightarrow \mathbb{R}$.
The author uses proof by contradiction, therefore assume there is a 1-1, onto function. This implies that no two "input" values from $\mathbb{N}$ will map us to the same value in $\mathbb{R}$ (1-1) and also that every element maps onto a unique value in $\mathbb{R}$, i.e. $x_1=f(1)$,$x_2=f(2)$ and so on, so we can write:
$$\mathbb{R} = \{x_1,x_2,x_3, ... \}$$
Would be the set of all the reals. Now the author proceeds to use the Nested Interval Property (Theorem 1.4.1 in the book) to produce a real that is not in the list.
Let $I_1$ be a closed interval that does not contain $x_1$ and $I_2$ be a closed interval that is contained in $I_1$ and which does not contain $x_2$. (1) My first understanding hurdle: "Certainly $I_1$ contains two smaller disjoint closed intervals", I cannot see which two disjoint intervals he has in mind. So I am starting to struggle at this point.
He then proceeds to say that $x_{n_0}$ is some real number from the list, then $x_{n_0} \notin I_{n_0}$ (fair enough, this is due to the way we constructed the intervals) and he then proceeds saying that:
$$x_{n_0} \notin \bigcap_{n=1}^{\infty}I_n$$
I do not see where the $I_{n_0}$ is in the above... The above has an intersection of the following intervals: $I_1,I_2,I_3,..$ and there is no $I_{n_0}$. 
(2) Finally, by construction there is a real that is not in the interval, $x_n \notin I_n$ is not in there at all. So by definition, we are omitting a real number from the interval, even though we are meant to prove it (I realize that this is an interval in the first place; My thinking was that if we prove it on the interval then we can extend the conclusions from that interval to the whole of the real line; But if we are omitting a real from the number line, then how can we make any conclusions at all).

REPLY [3 votes]: It is not important that every member of $\Bbb N$ be mapped to a unique real.  Somebody has given us a function that is claimed to be onto the reals, that is that every real is the image of at least one natural.  We are proving them wrong by finding a real they didn't cover.  If they cover some reals more than once, that is not a problem.  
For your first question, as an example, maybe $x_i=\pi$.  We now want to find a closed interval not including $\pi$ to call $I_1$.  It could be $[4,300]$, for example.  Now we are asked to find two closed intervals included within it.  We could take $[4,100]$ and $[101,300]$.  The particular intervals do not matter as long as they are disjoint.  Then at least one of them (maybe both) does not include $x_2$, so that one will become our $I_2$.  If $x_2=1$, it is in neither one and we can take either one as $I_2$.  If $x_2=10$ we are forced to take the second.  The point is just that we can find one.  Then we continue step by step.  
The claim about $x_{n_0}$really says that no real on the list is in $\bigcap_{n=1}^{\infty}I_n$.  $x_1$ isn't because it is not in $I_1$.  $x_{100}$ isn't because it isn't in $x_{100}$ and so on.  
In finally, we are not omitting a real from an interval, we are finding intervals that each do not cover a given real. Then we say the intersection contains a real, but we know it is not on the list.