The Setting
Let $K$ be an Archimedean field. TFAE:
- $K$ has the least upper bound property.
- Every Cauchy sequence in $K$'s additive group converges.
Now proving that 1 implies 2 is easy, but the other direction is slightly harder. Not that that's a problem. Rather the problem is that I can't see a route that doesn't invoke at least dependent choice at some point.
Strategy 1
Starting with a nonempty set $A$ that's bounded above, you could construct a monotonely non-decreasing Cauchy sequence of upper bounds that has the supremum as its limit. Here's a short sketch: Pick an upper bound $B_0$ of $A$. Pick an $a_0 \in A$. Recursively define $$ B_{i+1} = \begin{cases} \frac{B_i+a_i}{2}, & \text{ if that's an upper bound for } A \\ B_i, & \text{ otherwise} \end{cases} $$ and $$ a_{i+1} = \begin{cases} a_i, & \text{ if $\tfrac{a_i+B_i}{2}$ is an upper bound for $A$}\\ \text{choose any } a \in A \text{ s.t. } \frac{a_i+B_i}{2} < a, & \text{ otherwise.} \end{cases} $$ I can't see a way to get rid of the choice because you really want the $a_i$ to be in $A$ for the argument to go through.
Strategy 2
Okay, let's go the long way instead! First we show that $K$ complete implies $[a,b]$ compact. Then we show that that implies that closed and bounded subsets are compact ("Heine-Borel property"). And finally we show that not(Heine-Borel property) implies not(least upper bound property).
But I already get stumped on the first part. Clearly it's easy to show that $K$ complete implies $[a,b]$ is sequentially compact. And from here it'd be nice to use that $K$ is 2nd countable (the intervals with rational endpoints are a basis) to get that $[a,b]$ is in fact compact. So you start with an open covering $U_\alpha$ of $[a,b]$. 2nd countable spaces are Lindelöf... wait... let's make sure and prove that. Let $\{B_i\}$ be a countable basis. Then for each $B_i$ you choose a $U_\alpha$... oh. Choice crept up again.
So my question is this: Does this really require (an admittedly weak form of) choice? Or is there a way to do without?