In theorem $2$ of chapter $1$ of Real Mahtematical Analysis, by Chalrles C. Pugh, the author proves the existence of a least upper bound of any bounded subset $\mathbb{C}$ of $\mathbb{R}$ (which has been constructed by Dedekind's cuts already). In the proof, the author considers the cut $C\mid D$ where $C = \{a\in \mathbb{R}:a\in A$ for at least one $A\in \mathbb{C}\}$, $D=\mathbb{Q}-C$. Then he argues that $C$ is an upper bound for $\mathbb{C}$ because, (because of the construction), for any $A\in \mathbb{C}$, $A\subset C$. Then $C$ is proved to be the smallest upper bound by showing that any other upper bound must contain $C$ as well.
My confusion with this argument is that if we have used the fact that $\mathbb{C}$ is bounded above at all. Does it not follow from the aforementioned argument that any subset of $\mathbb{R}$ has an upper bound and has a least upper bound?
Your argument works up to a small, but essential, detail: in order that a set such as your $C$ is (the lower part of) a Dedekind cut, its complement $D=\mathbb{Q}\setminus C$ must not be empty.
Only if the subset $\mathcal{X}$ of $\mathbb{R}$ you start with is upper bounded you can conclude that, setting $$ C=\{a\in\mathbb{Q}: \text{$a\in A$, for some $A\in \mathcal{X}$}\} $$ is a properly defined Dedekind cut, in particular that $D=\mathbb{Q}\setminus C$ is not empty.
Note that $C$ must be a subset of $\mathbb{Q}$, not of $\mathbb{R}$ as you write.
Namely, if $r$ is an upper bound for $\mathcal{X}$, then $r$ is defined by a Dedekind cut $E\mid F$ and you should be able to prove that, for every $b\in F$, $b\notin C$.
Conversely, if $\mathcal{X}$ is not upper bounded, then you should be able to prove that $C=\mathbb{Q}$.