Let $A$ be a set of ordinals, I'd like to prove that $\bigcup A=\sup A$
The definitions I'm working with:
A set [of sets] $X$ is said to be transitive if $x \in X\implies x\subseteq X$
A set $X$ is said to be an ordinal if it is transitive and well ordered W.R.T. the relation $\in$
I'm stuck at showing $\bigcup A$ must be an upper bound of $A$. If we have $a\in A$, I'd like to show this implies $a\in \bigcup A$, but all I've arrived at is $a\subseteq A$ (I have already shown $A$ itself must be transitive) and $a\subseteq \bigcup A$ (by definition). How does this get me $a\in \bigcup A$?
$\bigcup A$ is transitive: If $y\in x\in \bigcup A$, then $x\in z$ for some $z\in A$ and by transitivity of $z$, $y\in z$, hence $y\in\bigcup A$.
$\bigcup A$ is well-ordered by $\in$: let $B\subseteq \bigcup A$ be non-empty, say $b\in B$. Then $b\in a$ for some $a\in A$. Then $a\cap B$ is a non-empty subset of $a$, hence has a minimal element $m$. Let $c\in B$. Then $c\notin m$ because $c\in m\in a$ would imply $c\in a\cap B$, contradicting minimality of $m$. As $m$ and $c$ are ordinals, this implies $m\in c\lor m=c$. Thus $m=\min B$.
Therefore $\bigcup A$ is an ordinal. So for $a\in A$, we have $a\subseteq \bigcup A$, and as $\bigcup A$ is an ordinal, this means $a\in\bigcup A\lor a=\bigcup A$.