 Solution

Fix a set $x$ and define by recursion on $\omega$The recursion theorem defines a function-class $F$ uniquely by $F(n)=T_n$: for each $n\in \omega$ there is exactly one finite sequence satisfying the displayed recursion through stage $n$. Thus $F$ is genuinely functional, and the Axiom schema of replacement applied to the set $\omega$ gives the set $\{T_n:n\in \omega \}$.

Now defineIt contains every member of $x$. If $y\in T$, then $y\in T_n$ for some $n$; hence every $z\in y$ belongs toThus $T$ is transitive. If $U$ is any transitive set containing every member of $x$, induction gives $T_n\subseteq U$ for every $n$, so $T\subseteq U$. Thereforeis the transitive closure of $x$.