Proving that $\mathbb{R}$ satisfies the Least Upper Bound property. If we state the Least Upper Bound property as an axiom for the real numbers, we can use this along with the other axioms for $\mathbb{R}$ to show that $\mathbb{R}$ is a complete metric space; so I would like to ask about the reverse argument:   
How can one show, using the assumption that $\mathbb{R}$ is a complete metric space (with the usual metric) and the ordered field axioms for $\mathbb{R}$, that $\mathbb{R}$ satisfies the Least Upper Bound property?
 A: You can do this as follows.
Let $A\subset\mathbb R$ be nonempty and bounded above. Choose $a_0\in A$ and an upper bound $b_0$ for $A$. Then, look at the middle point $m$ of the interval $[a_0,b_0]$. If $m$ is an upper bound for $A$, put $b_1:=m$ and $a_1:=a_0$. Otherwise, put $b_1:=b_0$ and choose a point $a_1\in A$ such that $a_1>m$. In either case, you have $a_0\leq a_1\leq b_1\leq b_0$, and $b_1-a_1\leq \frac12 \,( b_0-a_0)$. Moreover, $a_1\in A$ and $b_1$ is an upper bound for $A$.
Repeating this procedure, you can construct by induction a non-decreasing sequence $(a_n)\subset A$ and a non-increasing sequence $(b_n)$ of upper bounds for $A$, with $a_n\leq b_n$ for all $n$ and $b_{n+1}-a_{n+1}\leq \frac12 (b_n-a_n)$.
By the Archimedean property (which you do need as pointed out by Daniel), the diameter of the interval $[a_n,b_n]$ goes to $0$. It follows that both sequences $(a_n)$ and $(b_n)$ are Cauchy. So they are both convergent, to the same limit since $b_n-a_n\to 0$. If you call this limit $l$, then $l$ is an upper bound for $A$ because the set of all upper bounds for $A$ is closed and $b_n\to l$, and no upper bound for $A$ can be smaller than $l$ because $l$ is in the closure of $A$.
