If $S$ is a totally ordered field with the lub-property, you will not be able to prove constructively that any nonempty subset $H\subset S$ which is bounded from above has a least upper bound. This is a property you assume that $S$ has, not a property you prove that $S$ has.
It is, however, possible to prove that any nonempty subset $H\subset\bf R$ which is bounded from above has a least upper bound. The proof will depend on how you have defined the set of real numbers $\bf R$. Note that there are several definitions possible, all of which are equivalent. (The most common ones are by using Dedekind cuts of rational numbers, or by using equivalence classes of Cauchy sequences of rational numbers.)
If $S$ is a totally ordered field with the lub-property then there is a unique order-preserving field isomorphism $\phi : {\bf R}\to S$. This means that $\phi$ will have the following properties:
$$\phi \hbox{ is 1-1 and onto}$$ $$\phi(x+y)=\phi(x)+\phi(y)$$ $$\phi(xy)=\phi(x)\phi(y)$$ $${\rm If\ }x<y{\rm\ then\ }\phi(x)<\phi(y)$$
In order to prove that there exists such a map $\phi$, you follow the steps given in the answer by Carl Mummert. There is some work involved here, if you want to write out all details, but it is healthy work which will deepen your understanding of all the involved concepts.
You begin by showing that if there is such a map $\phi$, then it is necessary that $\phi(0)=0_S$ and $\phi(1)=1_S$, where $0_S$ and $1_S$ are the zero element and the unit element of the field $S$, respectively. Hence we define $\phi(0)=0_S$ and $\phi(1)=1_S$.
Now use the field axioms to define $\phi(x)$ for any rational $x$. Show that this can only be done in one way if $\phi$ is to be a field isomorphism. Show that $\phi$ will then be 1-1 on $\bf Q$. Also show that $\phi$ will preserve the ordering on rational numbers, so that if $x,y\in\bf Q$ and $x<y$ then $\phi(x)<\phi(y)$.
You now need to extend $\phi$ so that it becomes a map which is defined on all of $\bf R$, and not only on $\bf Q$. I'm sure it can be done in several (equivalent) ways, but the following seems natural: If $x\in\bf R$, let $E=\{y\in {\bf Q}: y<x\}$. Note that $E$ is bounded above in $\bf R$, and that $x$ is a least upper bound of $E$. Show that $\phi(E)$ is bounded above in $S$, and define $\phi(x)$ to be the least upper bound of $\phi(E)$. Show that this does not change $\phi(x)$ in the case when $x\in\bf Q$. Show that the extension still has the properties $\phi(x+y)=\phi(x)+\phi(y)$, $\phi(xy)=\phi(x)\phi(y)$, and that it still is order-preserving and 1-1.
You now have an order-preserving injective map $\phi:{\bf R}\to S$ which preserves the field structure. The final step is to show that it is necessarily surjective. Let $s\in S$, we need to find $x\in \bf R$ such that $\phi(x)=s$. First show that there are $a,b\in\bf R$ such that $\phi(a)<s<\phi(b)$. Define $E=\{y\in {\bf R} : \phi(y)<s\}$. Show that $E$ is bounded above, and let $x$ be the least upper bound of $E$. Show that $\phi(x)=s$.
Dedekind cuts are used for creating reals from rational numbers, that is, axiomatically, the reals are THE Dedekind cuts of the rationals. Without the condition, however, every rational would have two representations as a Dedekind cut: one where it is added to the lower class, and another in which it is added to the upper class. Hence the condition.
Best Answer
No.
A simple example that doesn't quite work is the total order on $\mathbb{R}\times\mathbb{R}$ defined such that $(a,b) < (c,d)$ if $a<c$ or if $a=c$ and $b<d$. (In other words, uncountably many copies of $\mathbb{R}$, lined up end-to-end.) This is dense and clearly not order-isomorphic to the real line (extended or not), because it contains uncountably many disjoint intervals that are order-isomorphic to the real line (unlike the real line itself, which can contain only countably many of these). However, it is not Dedekind-complete per your definition, because $\{0\}\times\mathbb{R}$ is bounded above (say, by $(1,0)$), but has no least upper bound. This can be repaired by considering $\mathbb{R}\times (\mathbb{R} + \{-\infty,\infty\})$ instead, in which case the supremum of $\{0\}\times \mathbb{R}$ is just $(0,\infty)$; and you are guaranteed greatest lower bounds as well.