2 Theorem
The set $\Bbb R$, constructed by means of Dedekind cuts, is complete in the sense that it satisfies the
Least Upper Bound Property:If $S$ is a nonempty subset of $\Bbb R$ and is bounded above then in $\Bbb R$ there exists a least upper bound for $S$.
Proof: Easy! Let $\mathcal C\subset\Bbb R$ be any nonempty collection of cuts which is bounded above, say by the cut $X | Y$. Define $C:=\{a\in\Bbb Q:\text{ for some cut } A | B\in\mathcal C\text{ we have } a\in A\}$ and $D=$ the rest of $\Bbb Q$.
source: Real Mathematical Analysis by Pugh.
I'm having trouble understanding what is meant by his definition of $C$, maybe an example would help? What would $C$ look like if the nonempty collection of cuts was $\{M|N, O|P, Q|R\}$? Is it saying that $C$ will take on all rationals from one of the cuts (say $M$) from the collection? Or will it have all rationals belonging to $M,O,$ and $Q$ i.e. $C=\{a\in M,O,Q \}$?
Though Robert Shore has answered your basic question, I thought a specific example might help. Let's do the classic: $\mathcal C = \{x \in \Bbb R : x < 0 \text{ or }x^2 < 2\}$. We can express this in terms of cuts in $\Bbb Q$ as $$\mathcal C = \{A|B \in \Bbb R : \text{ for all }a \in A, a < 0 \text{ or }a^2 < 2\}$$
Now for $a \in C$, there is some $A|B \in \mathcal C$ with $a \in A$. But by the definition of $\mathcal C$, this means that either $a < 0$ or $a^2 < 2$.
We note that the cut $X|Y = \{x \in \Bbb Q: x < 1\}|\{y \in \Bbb Q: y \ge 1\} \in \mathcal C$. Hence every $x < 1$ is in $X$, and therefore in $C$. Now suppose $r$ is any positive rational number with $r^2 < 2$. Then there must be a $q > r$ with $q^2 < 2$ (because there is a perfect rational square between any two rational numbers, which I'll leave to you to prove). But then $U|V = \{u \in \Bbb Q: u < q\}|\{v \in \Bbb Q: q \le v\}\in\mathcal C$, and $r \in U$. So $r \in C$.
Therefore $C = \{r \in \Bbb Q: r < 0 \text{ or } r^2 < 2\}$.
What is significant about this example is that there is no $A|B\in \mathcal C$ with $A = C$. $A \subset C$, but there is no single $A$ which is all of $C$.