Complete ordered field

You’re right: there is a problem with the argument, because there is a Cauchy-complete non-Archimedean ordered field, and a non-Archimedean ordered field is not complete in the sense of having least upper bounds.

The standard example starts with the field $F$ of rational functions over $\Bbb R$, with positive cone consisting of those functions $f/g$ such that the leading coefficients of $f$ and $g$ have the same algebraic sign. Then form the Cauchy completion by extending this to equivalence classes of Cauchy sequences in $F$. This is Example 7 on page 17 of Gelbaum & Olmsted, Counterexample in Analysis; you may be able to see it here.