Proof of Nested Interval Theorem: insufficient I have to prove that if $\mathrm{length}(I_{n}) \to 0$, then $\bigcap_{n=1}^{\infty} I_{n}$ contains precisely one point. 
The proof given in the text says that if $b_n>a_n$, $b_{n}$ converges to $b$  and $a_{n}$ converges to $a$ as $n\to\infty$, and $b_{n}-a_{n}\to 0$, then $a=b$.  
Convergence says that for any $\epsilon$, we can find $a_{n}$ and $b_{n}$ such that $|b_{n}-a_{n}|<\epsilon$. This is fine. It is however, also true that an infinite sequence can be made on every $\epsilon$- one of the many such sequences is $\epsilon, \frac {\epsilon}{2}, \frac {\epsilon} {2^2}\dots$
However small the value of $\epsilon$ may be, I don't see it ever containing 1 point. Reiterating, although we know $\epsilon\to0$, a length of $\epsilon$ will still contain an infinite number of points. 
Where am I going wrong? 
 A: So you are given a sequence of intervals $I_n:=[a_n,b_n]$ with $I_{n+1}\subset I_n$ for all $n\geq0$.
In order to proceed we have to agree on the fact that the system ${\mathbb R}$ is complete. This implies, e.g., that any nonempty and bounded set $S\subset{\mathbb R}$  has a least upper bound $\sup(S)\in{\mathbb R}$,  called supremum of $S$, and a largest lower bound $\inf(S)\in{\mathbb R}$, called infimum of $S$. (This is a statement about the fine structure of ${\mathbb R}$; it rules out holes in places where we hoped to find a number.)
The nesting of the intervals $I_n$ implies $$a_0\leq a_m\leq a_{\max\{m,n\}}\leq b_{\max\{m,n\}}\leq b_n\leq b_0\qquad \forall\ m,\ \forall n\tag{1}\ .$$ Therefore the set $A$ of left endpoints $a_n$ is bounded and therefore has a supremum $\alpha\in{\mathbb R}$, and similarly the set $B$ of right endpoints $b_n$ has an infimum $\beta\in{\mathbb R}$.
Consider a fixed $b_n$. By $(1)$ this $b_n$ is $\geq a_m$ for all $m$; therefore  $b_n$ is an upper bound for $A$, and this implies $b_n\geq\alpha$, since $\alpha$ is the smallest upper bound of $A$. Since this is true for each individual $b_n$ the number $\alpha$ is a lower bound of $B$, and $\beta$ being the largest such bound it follows that $\alpha\leq\beta$.
We now have $a_n\leq\alpha\leq\beta\leq b_n$ for all $n\geq0$, and this implies that any number $x$ with $\alpha\leq x\leq\beta$ is contained in all $I_n$. Assume that there are two such numbers $x<x'$. Then $[x,x']\subset I_n$ for all $n$, and this would imply that all $I_n$ have a length $\geq\delta:=x'-x>0$, contrary to assumption.
A: Perhaps it would be useful to test your intuition on an example.
Let the interval $I_n$ be $[0,\frac{1}{n}]$, so that $a_n=0$ for all $n$, and $b_n=\frac{1}{n}$ for all $n$. Then $a_n\to 0$ and $b_n\to 0$ as $n\to\infty$, and $|b_n-a_n|\to 0$, so the theorem applies. And we see that $\bigcap_{n=1}^\infty I_n=\{0\}$.
Do you agree that $\bigcap_{n=1}^\infty I_n=\{0\}$? Remember, given subsets $A_1,A_2,\ldots$ of $\mathbb{R}$, by definition we have
$$\bigcap_{n=1}^\infty A_n=\{x\in\mathbb{R}\mid x\in A_n\text{ for all }n\}.$$
Are there any numbers $x$, other than $0$, that are in the interval $[0,\frac{1}{n}]$ for every $n$?
A: Let (In)  0 and let A be the set of all points in ⋂∞ In. 

Assume A contains multiple points, a0,a1,a2… Then choose a0 such that |a0| < |ak| for all k ≠ 0, choose a1 such that |a1| < |ak| for all k ≠ 0, 1, etc. so that
|a0| < |a1| <|a2| < |a3| < …
But (In)  0 implies that for any ε > 0, ∃ n s.t. ∀x with |x|> ε, x does not lie in In.
Choosing ε = |a0|/2 implies that for some n,
ε = |a0|/2 < |a0| < |a1| <…<|ak|
and therefore ak does not lie in In for all k natural. 
This contradicts, so there cannot be multiple points in A.
Further, it is clear that no matter how ε > 0 is chosen, -ε < 0 < ε. This means if  (In)  0, then 0 lies in In for all n. Therefore 0 ∊ A, and combined with the previous result we conclude A = {0}.
A: Think you would be a big fan of non-standard analysis. In this form of calculus, the nested intervals contain points besides 0, and these points are defined in terms of a decreasing sequence. So the answer depends on the kinds of math you use.
A: Well , first of all $I$ = $\bigcap_{n=1}^\infty I_n $ can't contain two or more points : say I includes $x_1$ and $x_2$ then 
$ ||I||\leq||I_n||$   <  $|x_2-x_1|$ for sufficiently large n -- so we can't have both.
It contains precisely the unique point $x$ which is both the limit of $ a_n$ and $b_n$.
Since $\forall$n $a_n < x <b_n$ then $x \in I_n$ hence is an element of the infinite intersection by definition.
[Note that any finite intersection will have infinitely many points--but this is irrelevant ..]
We have our definition  :   $\bigcap_{n=1}^\infty A_n  =\{x\in\mathbb{R}\mid x\in A_n\text{ for all }n\}$
As an example --to test your understanding--  what would the intersection $\bigcap_{n=1}^\infty A_n  $  be  
if $A_n$=$[n,\infty)$ ??
A: The point I've tried to make in this thread is that there can be no intervals on $R$ containing only a finite number of points. However small the interval is, it still contains an infinite number of points. 
Someone answered that let's assume an interval containing two points $x_{1} and x_{2}$. This assumption itself is faulty, as there can be no such interval. If we take an interval $[x_{1},x_{2}]$, one of the many infinite sequences in it is  $x_{1}+\frac{x_{2}-x_{1}}{2}, x_{1}+\frac{x_{2}-x_{1}}{2^2}, \dots$. 
If we have an infinite decreasing sequence of intervals (in length), and we start out with an interval containing infinite points, we don't necessarily reach intervals with a finite number of points at some point in the sequence. 
