Let $\mathbb{F}$ be an arbitrary ordered field that satisfies Monotone Convergence Property then $\mathbb{F}$ is Order Complete.
Order Completeness Property:
If $S \subset \mathbb{F}$ is bounded above, then $\exists \ c \in \mathbb{F}$ that is an upperbound of S and for every upperbound $b$ of S, we have $c \leq b.$
Monotone Convergence Property
Every monotone bounded sequence in $\mathbb{F}$ is convergent in $\mathbb{F}$.
I am not posting anything new. The answers were given in parts here and here.
I have just brought it together for my own future reference.
Monotone Convergence Property $\implies$ Archimedean Property.
Let's assume $\mathbb{F}$ is not archimedean, $\exists$ $c \in \mathbb{F}$ with $n<c$ for all $n \in \mathbb{N}_{\mathbb{F}}$.
By assumption $(1,2,...)$ must converge,say to $r \in \mathbb{F}$. $\implies$ $(0,1,2,...)$ also converges to $r$.
Subtracting the two sequence we find (1,1,1,...) converges to 0, which is absurd.
Therefore $\mathbb{F}$ must be Archimedean.
Let $(\phi \ne)A \subseteq \mathbb{F}$ which is bounded above in $\mathbb{F}$, with $U$ being the set of upperbounds of $A$ in $\mathbb{F}$.
Want: To find an lub for $A$.
We now move on to build a few tools in order to prove the existence of $\operatorname{lub} A$.
Claim 1: $\{u-\varepsilon: u \in U\}=:U- \varepsilon \nsubseteq U$, for all $\varepsilon (>0) \in \mathbb{F}$.
Let $\varepsilon>0$.
Let $U-\varepsilon \subseteq U$, Now we use induction on $n \in \mathbb{N}_{\mathbb{F}}$.
If $U-n\varepsilon \subseteq U$ for some $n \in \mathbb{N}_{\mathbb{F}}$.
then $U-(n+1)\varepsilon=(U-\varepsilon)-n\varepsilon$ $\subseteq U-n\varepsilon$ $\subseteq U$. $\implies$ $U-n\varepsilon \subseteq U$ for all $n \in \mathbb{N}_{\mathbb{F}}$.
Hence by Archimedean Property we have $\mathbb{F}= \bigcup_{n=1}^\infty U-n\varepsilon \subseteq U$,
which contradicts $A \neq \phi$.
Claim 2: $\bigcap_{n=1}^\infty U-\frac{1}{n} \subseteq U$.
Let $x \in \bigcap_{n=1}^\infty U-\frac{1}{n}$ and let $(x<)y \in \mathbb{F}$.
By archimedean property $\exists$ $n \in \mathbb{N}_{\mathbb{F}}$ such that $x+\frac{1}{n} <y$.
$x \in U-\frac{1}{n}$\pause $\Rightarrow$ $x+\frac{1}{n} \in U$\pause $\Rightarrow$ $(x+\frac{1}{n}<)y \notin A$.
$y$ is chosen arbitrarily, so $x$ is an upperbound of A, i.e $x \in U$.
Claim 3: $U-\frac1n$ $\subseteq$ $U-\frac1m$ for all $m \le n$.
$m \le n \Rightarrow \frac1n \le \frac1m$.
$x \in U-\frac1n$ $\Rightarrow$ $x+\frac1n \le x+\frac1m$ $\Rightarrow x+\frac1m \in U$ $\Rightarrow x \in U-\frac1m$.
Corollary:
Let $n_k$ be any increasing sequence in $\mathbb{N}_{\mathbb{F}}$ then $\bigcap_{k=1}^\infty \left(U-\frac{1}{n_k}\right)$ = $\bigcap_{n=1}^\infty \left(U-\frac{1}{n}\right) \subseteq U.$
Claim 4: If $x \in \mathbb{F}$ with $x \notin U-\frac1n$, then $x<y$ for all $y \in U-\frac1n$.
Let $u-\frac1n<x$ for some $u \in U.$
Then $x-(u-\frac1n)>0$ $\implies u+x-(u-\frac1n) \in U$ $\implies x \in U-\frac1n$, contradiction.
Hence $x$ is less than every member of $U-\frac1n$.
Now we proceed in creating an increasing sequence so that we can apply Monotone Convergence Property.
Let $x_1\in (U-1) \setminus U$ ($\ne \phi$ by Claim 1)
$x_1\notin U$ $\Rightarrow$ $x_1 \notin \bigcap_{n=1}^\infty U-\frac{1}{n}$.(from Claim 2)
so $\exists$ $n_1>1$ such that $x_1 \notin U-\frac{1}{n_1}$.
Let $x_2 \in \left(U-\frac{1}{n_1}\right) \setminus U$.
Then $\exists$ $n_2>n_1$(from Claim 3)
such that $x_2 \notin U - \frac{1}{n_2}$.
Again consider $x_3 \in \left(U-\frac{1}{n_2} \setminus U\right)$ and so on.
This yields an increasing sequence $(1,n_1,n_2,...)$ in $\mathbb{N}_{\mathbb{F}}$(from Claim 3) and an increasing sequence $(x_k)$ in $\mathbb{F}$
(from Claim 4,3) such that
$x_k \in \left(U-\frac{1}{n_{k-1}}\right) \setminus U.$
Then $(x_k)$ is bounded above by each element of U.
$\implies$ $(x_k)\rightarrow x \in \mathbb{F}$.(M.C.P)
$\implies$ $x \le u$ for all $u \in U$.
Remains to show that $x \in U$
Let $x_k \leq x$ for all k and $x_k \in \left(U-\frac{1}{n_k}\right)$
$\implies x \in \bigcap_{k=1}^\infty \left(U-\frac{1}{n_k}\right) = \bigcap_{n=1}^\infty \left(U- \frac{1}{n}\right) \subseteq U$, i.e $x \in U$.
Therefore $x$ is the smallest element in $U$, which means $x=\operatorname{lub}A$.\
Hence for any nonempty bounded subset of $\mathbb{F}$, $\operatorname{lub} A \in \mathbb{F}$
i.e $\mathbb{F}$ is Order Complete.