[proofplan] We prove first that $\bigcup A$ is transitive by unpacking the definition of union and using transitivity of each ordinal in $A$. We then prove that membership well-orders $\bigcup A$: every nonempty subset has a membership-minimal element, obtained by first looking inside one ordinal containing a chosen element and then comparing ordinals. Finally, the two containment assertions follow directly from the definition of union and give precisely the supremum universal property. [/proofplan] [step:Prove that the union is transitive] Let \begin{align*} U := \bigcup A. \end{align*} We prove that $U$ is transitive. Let $x$ and $y$ be sets with $x \in y$ and $y \in U$. Since $y \in U$, by the definition of union there exists $\alpha \in A$ such that $y \in \alpha$. By hypothesis, $\alpha$ is an ordinal, hence $\alpha$ is transitive. Therefore $x \in y \in \alpha$ implies $x \in \alpha$. Since $\alpha \in A$, again by the definition of union, $x \in U$. Thus $U$ is transitive. [/step] [step:Show that membership well-orders the union] Let $B$ be a nonempty subset of $U$. Choose $b \in B$. Since $b \in U$, there exists an ordinal $\alpha \in A$ such that $b \in \alpha$. Hence $B \cap \alpha$ is nonempty. Because $\alpha$ is an ordinal, membership well-orders $\alpha$. Therefore there exists $m \in B \cap \alpha$ such that for every $z \in B \cap \alpha$ with $z \neq m$, one has $m \in z$. We claim that $m$ is membership-minimal in all of $B$. Let $c \in B$ with $c \neq m$. Since $c \in U$, there exists an ordinal $\beta \in A$ such that $c \in \beta$. Thus $c$ is an ordinal, because every element of an ordinal is an ordinal. If $c \in \alpha$, then $c \in B \cap \alpha$, and the defining property of $m$ gives $m \in c$. If $c \notin \alpha$, then $\alpha$ and $c$ are ordinals, so the trichotomy law for ordinals gives either $\alpha = c$ or $\alpha \in c$, since the alternative $c \in \alpha$ is excluded. If $\alpha = c$, then $m \in \alpha = c$. If $\alpha \in c$, then $m \in \alpha \in c$, and transitivity of the ordinal $c$ gives $m \in c$. Thus for every $c \in B$ with $c \neq m$, one has $m \in c$. Hence membership well-orders $U$. [/step] [step:Conclude that the union is an ordinal] An ordinal is a transitive set well-ordered by membership. The first step proves that $U = \bigcup A$ is transitive, and the second step proves that membership well-orders $U$. Therefore $\bigcup A$ is an ordinal. [/step] [step:Verify that every ordinal in the set is contained in the union] Let $\alpha \in A$. To prove $\alpha \subseteq U$, let $x \in \alpha$. Since $\alpha \in A$, the definition of union gives $x \in \bigcup A = U$. Hence $\alpha \subseteq \bigcup A$. [/step] [step:Verify the least upper bound property] Let $\gamma$ be an ordinal such that $\alpha \subseteq \gamma$ for every $\alpha \in A$. We prove $U \subseteq \gamma$. Let $x \in U$. By the definition of union, there exists $\alpha \in A$ such that $x \in \alpha$. By the assumed upper-bound property of $\gamma$, we have $\alpha \subseteq \gamma$, so $x \in \gamma$. Therefore $\bigcup A \subseteq \gamma$. Combining this with the previous step, $\bigcup A$ contains every element of $A$ and is contained in every ordinal upper bound of $A$. Thus $\bigcup A$ is the supremum of the set $A$ of ordinals. [/step]