Ok so firstly I can see how this makes sense. If you have say: $S=\{0,1,2\}$ with $0 = \emptyset, 1 = \{0\} = \{\emptyset\}, 2 = \{0,1\} = \{\emptyset,\{\emptyset\}\}$.
$$\cup S = \{\emptyset,\emptyset,\emptyset,\{\emptyset\}\} = \{\emptyset,\{\emptyset\}\} = \{0,1\} = 2$$
(I think this is right?)
Now I suppose I could prove it by induction but I was told of another proof where you show:
$\sup(S) \subseteq \cup S$ and then $\cup S \subseteq \sup(S) $ (Can someone explain how to write out this proof? I see how if you do both sides you are basically showing $\sup(S) = \cup S$)
Thanks!
Note that $\alpha\leq\beta$ if and only if $\alpha\subseteq\beta$ (either $\alpha\in\beta$, and it follows from transitivity, or $\alpha=\beta$).
Now if $\alpha\in S$ then $\alpha\subseteq\bigcup S$, so $\bigcup S\geq\alpha$ and therefore $\bigcup S\geq\sup S$.
On the other hand, if $\alpha<\bigcup S$ then there is some $\beta\in S$ and $\gamma\in\beta$ such that $\alpha<\gamma$. In particular $\alpha\subseteq\gamma\subseteq\beta$, so $\alpha<\sup S$. therefore $\bigcup S\leq\sup S$.
And equality follows.