Assume that every member of $\mathscr A$ is a transitive set.
(a) Show that $\bigcup \mathscr A$ is a transitive set.
(b) Show that $\bigcap \mathscr A$ is a transitive set (assuming that $\mathscr A$ is non-empty).
I'm going to prove it in MK's frame. My attempt:
(a) Consider
$$x\in\bigcup \mathscr A.$$
By definition, there exists $a\in A$ such that $x\in a$. Let now $z\in x$. Because $a$ is transitive, $z\in a$, so
$$z\in\bigcup \mathscr A$$
and thus $\bigcup\mathscr A$ is a transitive class. (To show that $\bigcup \mathscr A$ is a set we need to assume that $\mathscr A$ is also a set and apply the Axiom of amalgation).
(b) The proof is very similar: if
$$ x\in\bigcap \mathscr A $$
then $x\in a$ for every $a\in \mathscr A$. Now set $z\in x$. Since every $a\in \mathscr A$ is transitive,
$$ z\in a \qquad \forall a\in \mathscr A $$
and hence
$$ z\in \bigcap \mathscr A .$$
However, if $\mathscr A$ is non-empty, then we can assert that $\bigcap \mathscr A$ is a set.
Is the proof right?
I think the proof is correct, but it seems me very easy. I'm not very good at Set theory and this is an exercise form Enderton's book. I thought it was going to be more difficult.