Proof for any two ordered fields with the least upper bound property are order isomorphic I will write down the definitions first and then what I have done.
Order isomorphic:
$A$ and $B$ are ordered integral domains. They are order isomorphic if $\exists$ a bijection $f: A \to B$ such that
$$f(x+y) = f(x) + f(y)$$
$$f(xy) = f(x)f(y)$$
$$x < y \Rightarrow f(x) < f(y)$$
Let $A$ and $B$ be our ordered fields with least upper bound property.
Since they are fields, they are also integral domains.
Since they are fields, each one has an additive identity and a multiplicative identity. Let these four elements be $0_A, 1_A \in A$ and $0_B, 1_B \in B$
Any field that contains the integers contains the rationals as a subfield (I couldn't prove this). Also, I am assuming that this fields contain the integers, so they contain the rationals as well.
Can I set the following function? $f: A \to B$ such that
$$f(0_A) \to 0_B; f(1_A) \to 1_B; f(q1_A) = q1_B : q \in \mathbb{Q}$$
These function is linear, but I don't see where can I use the least upper bound properties of these two sets.
I don't know how to continue. Any help is appreciated! :)

There is another name for an ordered field that satisfies the least upper bound property -- the system of real numbers. So we are trying to construct an order isomorphism between any two systems of real numbers, say $\mathbb{R}_A$ and $\mathbb{R}_B$, in other words, the system of real numbers is unique. 
To extend the map $f(q1_A)=q1_B$, $q\in\mathbb{Q}$, for $r_A\in\mathbb{R}_A$, let $$S=\{q\in\mathbb{Q}|q1_A<r_A\}.$$
Then define $$f(r_A):=\sup\{q1_B|q\in S\}$$
where the supreme is taken in the system $\mathbb{R}_B$. 
Then you can try to check that this map $f$ indeed defines an order isomorphism between $\mathbb{R}_A$ and $\mathbb{R}_B$. 
