Archimedian field $K$ has LUB property iff it's complete requires DC? 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?
 A: It seems to me that you can modify strategy 1 as follows.  Given a nonempty set $A$ that's bounded above, let $x_0$ be a non-upper-bound for $A$ and let $y_0$ be an upper bound for $A$.  Recursively define
$$
y_{n+1} \;=\; \begin{cases}(x_n+y_n)/2 & \text{if this is an upper bound for $A$} \\
 y_n & \text{otherwise.}  \end{cases}
$$
and
$$
x_{n+1} \;=\; \begin{cases}(x_n+y_n)/2 & \text{if this is a non-upper-bound for $A$} \\
 x_n & \text{otherwise.}  \end{cases}
$$
Note that $x_1 \leq x_2 \leq \cdots \leq y_2 \leq y_1$. Furthermore, each $y_n$ is an upper bound for $A$, and each $x_n$ is a non-upper-bound for $A$.  It is easy to prove that both sequences are Cauchy sequences, and that they converge to the same limit $L$.  We claim that $L$ is the least upper bound for $A$.
Both directions are fairly easy.  If $a \in A$, then $a \leq y_n$ for all $n$, which proves that $a \leq L$.  Thus $L$ is in fact an upper bound for $A$.  Next, if $u$ is any upper bound for $A$, then $u \geq x_n$ for all $n$, and therefore $u \geq L$.  Thus $L$ is the least upper bound for $A$.
