Proof of the Monotone Convergence Theorem using Nested intervals Theorem? Nested Intervals Theorem: If  $I_{n}=\left [ a_{n},b_{n} \right ]$ and  $I_{1}\supseteq I_{2}\supseteq I_{3}\supseteq ...$ then $\bigcap_{n=1}^{\infty}I_{n}\neq \varnothing$ In addition if $b_{n}-a_{n}\rightarrow 0$ as $n \to \infty$ then  $\bigcap_{n=1}^{\infty}I_{n}$ consists of a single point 
Monotone Convergence  Theorem: If $a_{n}$ is a monotone and bounded sequence of real numbers then $a_{n}$ converges.
How can you prove the second theorem using only the first(WITHOUT using the least upper bound property, the Bolzano Weierstrass etc.)?
 A: It is not possible to prove that the Nested Interval Property implies the Monotone Convergence Theorem.  By that I mean that there are ordered fields with the Nested Interval Property that do not satisfy Monotone Convergence.
The examples I can think of involve non-standard models of analysis. For example, let $\mathbb{N}$ be the set of natural numbers, and let $D$ be a non-principal ultrafilter on $I$.  Then the ultrapower $\mathbb{R}^{\mathbb{N}}/D$ has the nested interval property but does not satisfy Monotone Convergence.   
Briefly, one constructs the ultrapower by first considering the product $\mathbb{R}^{\mathbb{N}}$, that is, the set of all sequences of reals. Two such sequences $(x_n)$ and $(y_n)$ are equivalent modulo $D$ is the set of $i$ such that $u_i=v_i$ is an element of the ultrafilter $D$.  On the ultrapower, one puts a ring structure by defining addition and multiplication coordinatewise modulo $D$. And if $(u_n)/D$ and $(v_n)/D$ are elements of $\mathbb{R}^{\mathbb{N}}/D$, we say that $(u_n)/D < (v_n)/D$ if the set of $n$ such that $u_n<v_n$ is an element of $D$. It turns out that the ultrapower just defined is a real-closed ordered field. The reals can be embedded in the ultrapower via equivalence classes of constant sequences. The ordering is non-Archimedean, since if $v_n=n$, then $(v_n)/D$ is larger than any $(u_{k,n})/D$, where $u_{k,n}$ is the integer $k$ for all $n$. 

If in addition to Nested Intervals, we ask that our field have the Archimedean Property, then Nested Intervals does imply Monotone Convergence. Let the sequence $(c_i)$ be say non-decreasing. For any $n$, let $a_n=c_n$. Let $b_1$ be an upper bound for the sequence $(c_i)$, and define $b_n$ as follows. If nothing below $b_{n-1}$ is an upper bound for $(c_i)$, let $b_n=b_{n-1}$. Otherwise, let $b_n=b_{n-1}-2^{k}$, where $k$ is the largest integer (possibly negative) such that $b_{n-1}-2^k$ is an upper bound for $(c_i)$.  From the Nested Interval Property for the sequence of intervals $(a_n,b_n)$, we can deduce the convergence of the sequence $(c_i)$.
Remark: There is a fair literature on the subject. A good survey, at least from the non-standard analysis side, can be found here.
