Please can you check my proof of the following?
The Nested Interval Property implies the Axiom of Completeness of the real numbers.
Nested interval property: If $I_1 \supseteq I_2 \supseteq I_3 \dots$ are closed intervals then $\bigcap_n I_n$ is not empty.
Axiom of completeness: If $S$ is a non-empty set in $\mathbb R$ that has an upper bound then $S$ has a least upper bound.
My proof: Let $S \subseteq \mathbb R$ be a set that has an upper bound $K$. The goal is to show that $S$ has a least upper bound. If $K$ is in $S$ then $K$ is a least upper bound of $S$. If $K$ is not in $S$ then there is $s \in S$ with $s < K$. Let $I_1 = [s,K]$. If there are no elements of $S$ between $s$ and $K$ then $s$ is a least upper bound of $S$. Otherwise let $s_2 \in (s,K]$ and $I_2 = [s_2,K]$. Proceed like this to either obtain a least upper bound of $S$ or infinitely many intervals $I_n$. If the process terminates with infinitely many intervals, by nested interval property the intersection $\bigcap_n I_n$ is non-empty. In this case, $x \in \bigcap_n I_n$ is a least upper bound of $S$. (proving this last claim is easy)
Best Answer
You say that proving the last claim is easy, but in fact that is where your problem is. (That's a good rule of thumb, by the way - if you are tempted to just say the proof is obvious, write it out anyway because it might not be!)
The trouble is that all you know is that $I = \displaystyle \bigcap_n I_n$ is nonempty - it may have more than one element, and for all you know, it might even contain elements of $S$! For instance, take $S = \{1-\frac{1}{n}\ |\ n\in \mathbb{N}\} \cup \{2-\frac{1}{n}\ |\ n\in \mathbb{N}\}$ and note that your procedure might end up with $I = [1, 3]$.