P Johnstone's "Notes on Logic and Set Theory" states that, for any set $x$, its transitive closure $\bigcup\{x\cup\bigcup x\cup\bigcup\bigcup x\cup\dots\}$ may be constructed using Union, Replacement, and Infinity.
How can this be proved? In particular, I don't see what definable function class to use in applying Replacement. (We can define informally that the ordinal 0 maps to $x$, 1 to $\bigcup x$, 2 to $\bigcup\bigcup x$, etc., but my understanding is that Replacement only applies if we can express the mapping as a function class, i.e. with a defined predicate.)
I don't want to use the Recursion Theorem, because the text uses transitive closure along the way in proving the recursion theorem.
Define the following function $F(n,x)=y$ if and only if there exists a sequence $x_i$ for $i\leq n$ with $x_0=x$ and $x_n=y$, such that for all $i<n$, $x_{i+1}=\bigcup x_i$.
Prove that this is a function, and that it is well-defined for all $n<\omega$. Then appeal to Infinity, Replacement, and finally Union.
Another approach to do that is to appeal to the von Neumann hierarchy. Once you know that $x$ is a subset of some transitive set, you can just take the intersection of all transitive sets containing $x$ and prove that it is indeed the transitive closure of $x$.
If you're unhappy with applying the Recursion Theorem (for $\omega$), then you're likely to be unhappy with this solution either, as it requires transfinite recursion to define the von Neumann hierarchy.