I will show about the strategy of this proof. We want to show that
The existence of a least upper bound on bounded sets by above imply that...
...if we have have two sets $X$ and $Y$ such that $x\le y$ for all $x\in X$ and all $y\in Y$ then exists a $c$ such that $x\le c\le y$ for all $x\in X$ and all $y\in Y$
Then we want show that for $S:=\sup(X)$ holds
$$x\le S\le y,\quad\forall x\in X,\forall y\in Y\tag{1}$$ provided that $x\le y$ for all $x\in X$ and all $y\in Y$.
Because $S$ is the supremum of $X$ then the LHS of (1) is clear, that is, $x\le S$ for all $x\in X$ by the definition of supremum. Then to conclude the proof we must show that $S\le y$ for all $y\in Y$.
Then we show that if $S\le y,\forall y\in Y$ is not true then it must be the case that exists some $y_0\in Y$ such that $y_0<S$. But then this would imply that $S$ is not the supremum of $X$, that is, $x\le y_0<S$ for all $x\in X$, a contradiction.
Hence if $S$ is the supremum of $X$ it must be the case that $S$ is a lower bound of $Y$, that is, that $x\le S\le y$ for all $x\in X$ and for all $y\in Y$.$\Box$
We can use the Cantor-intersection theorem and the Archimedean axiom to prove the Dedekind's theorem (source:Enrico Gregorio's answer).
Suppose $A,B\subseteq\Bbb R$ are non-empty s. t. $(\forall a\in A)(\forall b\in B)\quad a\le B$ and $A\cup B=\Bbb R$. Then $\exists ! c\in\Bbb R$ s. t. $a\le c\le b, \forall a\in A,\forall b\in B$.
Let's abbreviate the above with $A\le B$ and $A\le c\le B$.
Since $A\ne\emptyset\space\land\space B\ne\emptyset$, let's choose $a_0\in A$ and $b_0\in B$.
Now, we choose $c_0=\frac{a_0+b_0}2$.
If $A\le c_0\le B$, we're done. Otherwise, it is either
$$c_0<a,\text{for some }a\in A\space\text {or}\space c_0>b\text{ for some }b\in B.$$
If $c_0<a$, then let $a_1=a, b_1=b_0$.
If $c_0>b$, then let $a_1=a_0, b_1=b$.
In either case, $a_0\le a_1\space\land\space b_0\ge b_1\implies [a_0,b_0]\supseteq[a_1,b_1]$.
Let $d=b_0-a_0$. We have $b_0-c_0=c_0-a_0=\frac{b_0-a_0}2=\frac{d}2$.
If $c_0<a$, then $\frac{d}2=b_0-c_0>b_0-a=b_1-a_1$.
If $c_0>b$, then $\frac{d}2=c_0-a_0>b-a_0=b_1-a_1$.
We can repeat the procedure to build a sequence of nested closed intervals:
$$[a_0,b_0]\supseteq[a_1,b_1]\supseteq\cdots\supseteq[a_n,b_n]$$
If at some point $A\le c_n\le B$, we're done. Otherwise, we get a sequence of closed nested intervals $[a_n,b_n]$ with $a_n\in A,b_n\in B$ and $b_n-a_n\le\frac{d}{2^n}.$
By the Cantor-intersection theorem (now assumed as an axiom), there is $c\in\bigcap_n[a_n,b_n]$.
Now we want to prove $A\le c\le B$. Let's assume the opposite: there is either
$$a\in A\text{ with } c<a\space\text{ or }\space b\in B\text{ with } c>b.$$
Suppose $a>c$.
Let's take $\varepsilon=\frac{a-c}2$.
We can take $n\in\Bbb N$ s. t. $\frac{d}{2^n}<\varepsilon$. $(*)$
Then we obtain:
$b_n<a+\varepsilon\le c+\varepsilon=c+\frac{a-c}2=\frac{a+c}2\le a$ which contradicts $A\le B$.
Analogously in case of $c>b$.
$(*)$ As $2^n>n, n\in\Bbb N$, we have $\frac1{2^n}<\frac1n\implies\frac{d}{2^n}<\frac{d}n$. In order to make $\frac{d}{2^n}<\varepsilon$, we could take $n$ s. t. $\frac{d}n<\varepsilon$, which is guaranteed to hold true by the Archimedean axiom.
Now, we want to prove the Dedekind's theorem is equivalent to the axiom of completeness (like Henno Brandsma did here):
Every non-empty subset of $\Bbb R$ that is bounded above has a supremum in $\Bbb R$.
Direction $\boxed{\Rightarrow}$
Suppose Dedekind's theorem holds.
Let $S\subset\Bbb R$ with some upper bound $u$ and let's define:
$$A=\{x\in\Bbb R\mid\exists s\in S, x<s\}\\\text{and}\\ B=\{x\in\Bbb R\mid\forall s\in S, x\ge s\}$$
$B\ne\emptyset$ because $u\in B$.
Then our $c=\sup S$.
Direction $\boxed{\Leftarrow}$
Assume the axiom of completeness holds.
Then we take $c=\sup A$.
Best Answer
I've never heard anyone state the axiom of completeness as "there exists a set $R$ such that...". Phrasing it this way makes it sound like an axiom of set theory. Usually the axiom of completeness is an axiom about an ordered Archimedean field, i.e., it just says "any bounded nonempty subset of $R$ has a least upper bound". And then it's a theorem that an ordered Archimedian field satisfying this axiom exists.
Also, I have to nitpick about the way you asked the question: both of the statements you write are provable (say in ZF set theory), so yes, they are equivalent over ZF set theory. On the other hand, they are certainly not logically equivalent (i.e., equivalent on the basis of no axioms).
I think the proper way to frame the question is as follows: Let $R$ be an ordered Archimedean field. Is it true that $R$ is complete (any bounded subset of $R$ has a least upper bound) if and only if $R$ is maximal ($R$ is not a proper subfield of an ordered Archimedean field). The answer is yes. I'll sketch the proof.
In one direction, suppose $R$ is complete. Let $R\subseteq R'$, where $R'$ is an ordered Archimedian field. Let $x\in R'$, and let $L_x = \{r\in R\mid r<x\}$. Since $R'$ is Archimedian, $L_x$ is non-empty and bounded, so it has a least upper bound $y\in R$. If $x\neq y$, then $y-x$ is a positive infinitesimal, contradicting the assumption that $R'$ is Archimedean. Thus $x = y\in R$, so $R' = R$. Thus $R$ is maximal.
Conversely, suppose $R$ is maximal. Construct the Dedekind completion $R'$ of $R$. By maximality, $R' = R$, so $R$ is complete.
On the "dual version": Yes, $\mathbb{Q}$ is the minimal ordered Archimedean field in the sense that no proper subfield of $\mathbb{Q}$ is an ordered Archimedean field. But writing "ordered Archimedean" twice is overkill here. $\mathbb{Q}$ is a minimal field in the sense that has no proper subfields at all. You can just as well characterize $\mathbb{Q}$ as the unique minimal field of characteristic $0$ up to isomorphism.