The axiom of regularity in a fact about $V_\omega$ I'm working out the details  in this proof (from here), and I have some questions about the second part ("For the other direction, ...").
First, why is it possible to assume WLOG that $A$ is transitive?
Also, it says "Now take a minimal element $x_0$ in $B$ by the Axiom of Regularity". The axiom of regularity says that in any non-empty set, there is an element that doesn'h have elements in common with that set. I can see how this axiom is used in proving that $x_0\subseteq V_\omega$. But why is the author mentioning "a minimal element"? In particular, minimal in what sense? Is it some other form of the axiom of regularity?
My last question is not about the proof but rather about the statement of the theorem being proved. In another textbook, I found the following statement: "$x\in V_\omega$ iff $x$ is well-founded and $trcl(x)$ is finite", and there's a remark saying that the condition that $x$ be well-founded cannot be omitted. I believe the statement "$trcl(x)$ is finite" is equivalent to "$x$ is hereditarily finite", and if so, the theorem being proved says "$x\in V_\omega$ iff $trcl(x)$ is finite", omitting the condition that $x$ be well-founded. Is well-foundedness still used implicitly in the proof (maybe when applying the axiom of regularity), and should it be included in the statement of the theorem?
 A: Note that in $\mathsf{ZF}$, every set is well-founded, as well-foundedness of a set $x$ means that $\in$ is a well-founded relation on $\mathrm{tc}(x)$, which is true by the axiom of regularity.
The term minimal in the author's argument means exactly what you are writing, note that an element $y \in x$ with $y \cap x = \emptyset$ is minimal with respect to the relation $\in$ on $x$.
For your first question: If $A$ is not transitive, consider $A' = \mathrm{tc}(A)$. Then $A'$ is hereditarily finite and $A' \in V_\omega \iff A \in V_\omega$.
A: If $A$ is hereditarily finite but $A\not\in V_{\omega}=\cup_{n\in \omega}P(V_n)$
then $\forall n\in\omega\,(A\not\subset V_n)$
so $\forall n\in\omega\,(trcl(A)\not\subset V_n)$... (because $trcl(A)\supset A$)...
so $trcl(A)\not\in \cup_{n\in\omega}P(V_n)= V_{\omega}.$
So if  A is hereditarily finite but $A\not\in V_{\omega}$ then there exists a  hereditarily finite transitive set (namely,$trcl(A)\,)$ that is not in $V_{\omega}$. This explains  "WLOG $A$ is transitive" because the book gives a proof that a hereditarily finite transitive set  that is not in $V_{\omega}$ is impossible.
