Ordered Field with Archimedean Property and Convergent Cauchy Sequences satisfies Completeness How would I start with the following problem?

Show that if $\mathbb{F}$ is an ordered field such that
(a) the order of $\mathbb{F}$ satisfies the Archimedean Property, and
(b) every Cauchy sequence is convergent,
then $\mathbb{F}$ satisfies the Completeness Axiom.

I'm stuck and I'm not sure where to begin. Would a proof by contradiction work well?
 A: You begin by noting that the completeness axiom states the existence of a specific element, namely the supremum of a given bounded set. Therefore the first thing to look at is whether any of the provided conditions also states the existence of an element. Actually both do, but the Archimedean property only states the existence of a natural number (namely a natural number larger than a given field element; note that I here identify the natural numbers with the corresponding multiples of the field's multiplicative identity), while the convergence of Cauchy sequences doesn't have a limitation on the type of number (other than it has to be a member of $F$, of course).
This suggests the strategy of constructing a Cauchy sequence whose limit is necessarily the supremum of the given set $S$. Since the Cauchy sequence is known to converge by assumption, this then implies that the supremum exists.
One obvious strategy is to construct a decreasing Cauchy sequence of upper bounds. You start with an arbitrary upper bound $b$ of $S$, and then go downwards approaching the set from above. But how much to go downwards? After all, if we don't go forward sufficiently far, we may end at an upper bound of $S$ that is not the supremum; maybe even far away from it.
But what we know for sure is that if we reach any point of $S$ other than the possible maximum, we've gone too far (that is, the sequence no longer consists of upper bounds). Note that if $a$ happens to be the maximum of $S$, we already know that it has to have a supremum, namely $a$ itself; therefore we can for the moment simply ignore that case. So select an arbitrary $a\in S$ as "way mark" to which we want to get as close as possible, but no closer.
Now thanks to our experience with real numbers (or alternatively remembering Zeno's paradoxes), we can guess the following strategy:
Since we know that we cannot go the full way to $a$, try to go just half the way. Since we start with an impossible way (namely from $b$ all the way to $a$), and going twice the half way is equivalent to going the full way, this means that in each step we consistently try half the step size from the previous step.
Thus we can recursively define the sequence definition as follows:
$$x_0 = b; x_{n+1} = \begin{cases}
x_n - \frac{b-a}{2^n} & \text{if this happens to be an upper bound of $S$}\\
x_n & \text{otherwise}
\end{cases}$$
This is clearly a (non-strictly) decreasing sequence of upper bounds. But is it a Cauchy sequence?
Well, what we can immediately see is that $\left|x_{n+1}-x_n\right|\le\frac{b-a}{2^n}$, which itself is a decreasing sequence. So for any $\epsilon>0$, if we find one value of $n$ such that $\frac{b-a}{2^n}<\epsilon$, then it will be the case also for all larger values of $n$.
Therefore what we need to prove is that for any $\epsilon>0$ there exists an $n\in\mathbb N$ so that $\frac{b-a}{2^n}<\epsilon$, that is, $2^n>\frac{b-a}{\epsilon}$. Now the left hand is just arithmetics of natural numbers; from that we know that $2^n>n$ for all $n\in\mathbb N$. So if we can find an $n$ larger than $\frac{b-a}{\epsilon}$, then $2^n$ will be larger, too.
But that's exactly what the Archimedean property tells us: For any number in $F$, there's a natural number $n$ such that $n>F$. Thus the Archimedean property tells us that this sequence we constructed indeed a Cauchy sequence, and therefore is has a limit. Let us therefore define
$$s := \lim_{n\to\infty} x_n$$
But is this $s$ we just defined really the supremum of $S$? Well, to start with, it is clearly an upper bound: If there were an $c\in S$ with $c>s$, then by convergence of $x_n$ to $s$ we would find an $x_n$ with $x_n-s = |x_n-s| < c-s$, that is, $x_n < c$. But $x_n$ is by construction an upper bound, so this cannot happen.
On the other hand, assume there's an upper bound $u<s$. Then obviously for all $x_n$ we have $x_n-u  > s-u > 0$. Now again using the Archimedean property we can deduce that there exists an $N$ such that $s-u > \frac{b-a}{2^n}$. Clearly that means that for any $n\ge N$, $x_n - \frac{b-a}{2^n}$ is an upper bound, and thus by the definition of the sequence is the value of $x_{n+1}$. Therefore either $x_n<x_{n+1}$ for all $n\in\mathbb N$, or there must exist an $m\in \mathbb N$ such that $x_m = x_{m+1}$ but $x_n\ne x_{n+1}$ for all $n>m$. However in that case one can show that $s=x_m + \frac{b-a}{2^n}$, which therefore is an upper bound. But then, by the definition of the sequence, we should have had $x_{m+1} = x_m + \frac{b-a}{2^n}$, in contradiction to $x_m=x_{m+1}$. So the only remaining case is $x_{n+1}<x_n$ for all $n\in\mathbb N$. But in that case, $s=a\in S$, and thus there cannot be a lower bound below $s$.
Note that the last case exactly means that the $a\in S$ we chose is indeed the maximal element. So this case is actually included in that construction, although it was explicitly not considered when constructing it.
