Does the proof that the Nested Set Theorem implies the Axiom of Completeness require the Archemedean property? Here is the beginning of a standard proof that the Nested Set Theorem implies the Axiom of Completeness (e.g. see here and here):
Suppose $E$ is a non-empty set bounded above by $b$ and let $a$ be an element of $E$.
    Define the following sequence of sets:
    \begin{gather*}
    [a_1, b_1] = [a,b] \\[8pt]
    [a_{n+1}, b_{n+1}] = \begin{cases}
\left[a_n, \dfrac{a_n + b_n}{2}\right] & \text{if $\, \dfrac{a_n + b_n}{2}\, $ is an upper bound of $E$} \\[8pt]
            \left[\dfrac{a_n + b_n}{2}, b_n\right]& \text{otherwise}
\end{cases}
\end{gather*}
    Then $[a_1, b_1]$ is bounded and $\{[a_n, b_n]\}_{n=1}^\infty$ is a descending countable collection of nonempty closed sets of real numbers, so the Nested Set Theorem implies that the intersection $\bigcap_{n=1}^\infty [a_n,b_n]$ is non-empty.
    Let $x$ be an element of this intersection.
Correct me if I'm wrong, but to show that $x$ is the least upper bound of $E$ you need to know that $b_n \to x$ and $a_n \to x$. Is it possible to know this without invoking the Archemedean property of the reals? 

You did not say what exactly you mean by nested set theorem, but I will assume that you mean the nested interval property:

If $\{F_n\}$ is a sequence of closed bounded intervals such that 
  $F_{n+1} \subseteq {F_n}$ for every $n \in \mathbb{N}$, 
  then $\bigcap_{n=1}^\infty \ne \emptyset$.

This, combined with the ordered field axioms, does not imply order-completeness. A useful reference for this type of question is James Propp's
Real analysis in reverse. On page 13 (of version 4) it says

The Nested Interval Property (18) does not imply completeness

and gives a number of references for counterexamples.
So you do need some extra hypothesis, such as the Archimedean property.
