Let $R$ be an arbitrary Ordered Field.
Order Completeness Property: If $S \subset R$ is bounded above, then $\exists \ c \in R$ that is an upperbound of S and for every upperbound $b$ of S, we have $c \leq b.$
Nested Interval Property: If $I_1,I_2,...,I_n,...$ be a collection of nested closed intervals, i.e $I_1 \supseteq I_2 \supseteq ... \supseteq I_n \supseteq ...$ then $\bigcap_{1}^{\infty} I_i \neq \phi$.
Order Completeness $\implies$ Nested Interval Property.
This part is easy to prove.
Nested Interval Property $\implies$ Order Completeness.
Consider $A \underset{bdd.}\subseteq R$,
$b_0$: an upperbound of $A$,
$a_0$: not an upperbound of $A$.
$I_0:=[a_0,b_0] \implies I_0 \cap A \neq \phi$.
Let $c=(a_0+b_0)/2_R$, If $\exists \ a \in A$ with $c<a$ then choose $a_1=c$, $b_1=b_0$.
Otherwise choose $a_1=a_0$, $b_1=c$.
Proceeding this way one can construct a sequence of
closed nested intervals $I_0 \supseteq I_1 \supseteq ... \supseteq I_n \supseteq ...$, in
which
$I_n=[a_n,b_n]$ where
$a_n$: not an upperbound of A,
$b_n$: upperbound of A.
Now $\operatorname{diam}(I_n)=\frac{b_0-a_0}{2^n}$. Assuming R enjoys Archemedian Property I can conclude $\operatorname{diam}(I_n) \to 0$
$\implies$ $\bigcap_{1}^{\infty} I_i=\{s\}$.
Remains to show that $s=\operatorname{lub}A$.
If there exists $a \in A$ such that $s<a$ since $\operatorname{diam}(I_n) \to 0$ there exists $p \in \mathbb{N}$ such that $\operatorname{diam}(I_p)<a-s$. Now $s \in I_p$ and $a \notin I_p \implies b_p<a$, contradiction to the fact that $b_p$ is an upperbound of $A$.
Hence $s$ is an upperbound of A.
Again let $l<s$ then there exists $q \in \mathbb{N}$ such that $\operatorname{diam}(I_q)<s-l$.
We have $s \in I_q$ then $l \notin I_q$ $\implies l<a_q$. Now there exists $b \in A$ with $l<a_q<b \implies l$ is not an upperbound of A.
Hence there exists $s \in R$ such that $s=\operatorname{lub}A$.
Now my question is If I drop the assumption of Archimedian Property can we still conclude this result?
Certain ordered fields $R$ are $\mathbf{\aleph_1}$-saturated as linear orders. This means that given countable subsets $A,B\subset R$ with $A<B$, there is $x \in R$ with $A<x<B$. Such an ordered field is not archimedean, since there is $x \in R$ with $\mathbb{N}<x<\varnothing$.
But such a field satisfies the nested interval property because writing $I_n=[a_n,b_n]$ for each $n \in \mathbb{N}$ and considering $x \in R$ with $\{a_n: n \in \mathbb{N}\}<x<\{b_n: n \in \mathbb{N}\}$, one has $x \in \bigcap \limits_{n \in \mathbb{N}} I_n$ (in the non-degenerate case where $(a_n)_{n \in \mathbb{N}}$ or $(b_n)_{n \in \mathbb{N}}$ is not stationnary).
So on its own the nested interval property does not imply completeness.
An example of an $\aleph_1$-saturated ordered field is the field $\mathbf{No}(\omega_1)$ of surreal numbers of countable length.