[proofplan] We verify the least-upper-bound property directly in the partially ordered set $(\mathcal{P}(X), \subset)$. First we show that the indexed union is itself an element of $\mathcal{P}(X)$. Then we prove that it contains every member of $\mathcal{A}$, and finally that every other upper bound contains it. [/proofplan]

[step:Define the candidate supremum as a subset of $X$] Define \begin{align*} U := \bigcup_{B \in \mathcal{A}} B. \end{align*} We first check that $U \in \mathcal{P}(X)$. Let $x \in U$. By the definition of union over a family of sets, there exists $B \in \mathcal{A}$ such that $x \in B$. Since $\mathcal{A} \subset \mathcal{P}(X)$, every $B \in \mathcal{A}$ satisfies $B \subset X$, so $x \in X$. Hence $U \subset X$, and therefore $U \in \mathcal{P}(X)$. [/step]

[step:Show that the union is an upper bound for the family]Let $B \in \mathcal{A}$. We prove $B \subset U$. If $x \in B$, then, because $B$ is one of the sets indexed in the union defining $U$, the definition of union gives $x \in U$. Therefore $B \subset U$ for every $B \in \mathcal{A}$, so $U$ is an upper bound for $\mathcal{A}$ in $(\mathcal{P}(X), \subset)$.[/step]

[guided]To prove that $U$ is an upper bound, we must use the order relation in the ambient partially ordered set. Here the order relation is subset inclusion, so the upper-bound condition says precisely: \begin{align*} B \subset U \end{align*} for every $B \in \mathcal{A}$. Fix an arbitrary $B \in \mathcal{A}$. To prove $B \subset U$, take an arbitrary element $x \in B$. Since $U$ was defined by \begin{align*} U := \bigcup_{C \in \mathcal{A}} C, \end{align*} and since the particular set $B$ belongs to the indexing family $\mathcal{A}$, the membership condition for a union of a family gives $x \in U$. Thus every element of $B$ is an element of $U$, so $B \subset U$. Because $B \in \mathcal{A}$ was arbitrary, $U$ is an upper bound for the whole family $\mathcal{A}$.[/guided]

[step:Show that every upper bound contains the union] Let $C \in \mathcal{P}(X)$ be an upper bound for $\mathcal{A}$ in $(\mathcal{P}(X), \subset)$. By the definition of upper bound, for every $B \in \mathcal{A}$ one has $B \subset C$. We prove $U \subset C$. Let $x \in U$. By the definition of $U$, there exists $B \in \mathcal{A}$ such that $x \in B$. Since $C$ is an upper bound for $\mathcal{A}$, we have $B \subset C$, and hence $x \in C$. Therefore $U \subset C$. [/step]

[step:Conclude the least-upper-bound property] We have shown that $U$ is an upper bound for $\mathcal{A}$ and that $U \subset C$ for every upper bound $C \in \mathcal{P}(X)$ of $\mathcal{A}$. Therefore $U$ is the least upper bound of $\mathcal{A}$ in $(\mathcal{P}(X), \subset)$. Hence \begin{align*} \sup \mathcal{A} = U = \bigcup_{B \in \mathcal{A}} B. \end{align*} If $\mathcal{A} = \varnothing$, the same argument gives $U = \varnothing$, and $\varnothing$ is contained in every element of $\mathcal{P}(X)$, so the conclusion also covers the empty-family case. [/step]