Dedekind Cuts versus Cauchy Sequences

They are the same (up to order-preserving isomorphism). I defer to Theorem 15 of a previous answer of mine.

Moreover, your first definition is wrong. It's supposed to be equivalence classes of rational Cauchy sequences - otherwise you have a ring but not a field.

In the real numbers case, the construction is isomorphic, as kahen said in his answer.

If you look at this fact closely enough, you realize that this is because the backbone of the real line is actually the integers, which has no accumulation points within itself, that is if a sequence converges it has to be constant from some point.

This fact, while seems unrelated, helps us avoid the following case: Consider some dense linear order (like the rationals) only bigger. For example, put two copies of the rationals one on top of the other, of course this can be embedded back into the rationals themselves, but consider the Dedekind closure of this order. It would have a point which is the exact point in which you switch from the first copy of the rationals to the second one. (This might not be such a good example, if you have a better example - this post is CW for this very reason)

So now let us try and prove something (together, I got stuck halfway through and although the intuition is clear to me, I'm uncertain how to put it into words - and this might as well be complete rubbish too.)

Theorem: Suppse $F$ is an ordered field (so it has characteristics 0, and we can assume $\mathbb{Q}\subseteq F$ without the loss of generality) such that its Cauchy completion (denoted by $C(F)$) and Dedekind completion (denoted by $D(F)$), both form a field - and it is the same field (namely there is a ring isomorphism between them).
Then $F\subseteq\mathbb{R}$.

Proof: Suppose not. Since $\mathbb{Q}\subseteq F$ then $\mathbb{R}\subseteq C(F)$ (and from our assumption, $\mathbb{R}\subseteq D(F)$ as well - they are isomorphic). Now let $a\in C(F)\setminus\mathbb{R}$ (without loss of generality take $a>0$), that is it is not a Cauchy limit of rational sequences. Meaning every $a_n\to a$ has only finitely many rational elements, so again without the loss of generality $a_n$ is not rational for all $n$. Since $D(F)$ is an ordered field so is $C(F)$, if there was $r\in\mathbb{R}$ such that $a<r$ then we have that $a\in (0,r)$ and so we can find a rational Cauchy sequence that converges to $a$. That to say that $a>r$ for any real number $r$.

Once more, we use the isomorphism, $a$ defines a cut which means the set $A = \{x\in F| x>q,\ \forall q\in\mathbb{Q}\}$ is non-empty, take $z$ to be the cut defined as the number between $\mathbb{Q}$ and $A$ (usually denoted by $z=\langle \mathbb{Q}\big| A\rangle$). Take some $\epsilon>0$ then $z-\epsilon$ is already smaller than some rational number, and therefore can be expressed as a limit of rational numbers, ergo $z-\epsilon\in\mathbb{R}$.

Take $z_n \to z$ some Cauchy sequence, since $z$ cannot be expressed as a real number, we can assume $z_n$ is not rational for all $z$, and therefore even if we take real numbers we cannot express $z$ as a Cauchy limit of such real numbers. However, for all $\epsilon>0$ we can express $z-\epsilon$ as a Cauchy limit of real numbers, so we can choose some sequence that would converge to $z$ Cauchy wise, which is a contradiction.

I am quite certain that I made a big mess in this proof, and would like it very much if someone will clean it up a bit. It should probably be easier to prove it by some model theoretic method, or maybe by using equivalent topologies of order and metric. I'm not sure.

That being said, I completely resent the whole idea that we based metric spaces on the real line and then we say there are no metric-complete ordered fields except $\mathbb{R}$.