Let $D$ be a transitive set with the following property : $$\forall a\in D:\quad a\subseteq B\,\Longrightarrow a\in B$$ Prove that $D\subseteq B$.
I think there will be needed axiom of regularity in my attempt, but don't know how to apply it :
By contrary, suppose $a_0\in D$, but $a_0\notin B$. Therefore $a_0\nsubseteq B$. Thus, there exists $a_1\in a_0\in D$, such that $a_1\notin B$. By repeating this procedure $\alpha$ times, Now look at $D-B$ :
$$\left\{\begin{array}{ll} a_{\alpha}\in\cdots\in a_2\in a_1\in a_0\in D-B\\ \forall\alpha:a_{\alpha}\in D-B\quad(\text{by transfinite induction!}) \end{array}\right.$$
Now the rank of sequence $\{a_{\alpha}\}_{\alpha\in Ord}$ is decreasing without end (But it's not precise anyway!).
Any help for formulation and making precise is so appreciated. (Or any approach other than regularity)
Instead of doing transfinite induction directly, just use the (already known, I hope) fact that all sets have rank:
Suppose $D\not\subseteq B$. Then $D\setminus B$ is nonempty and therefore has an element $x$ of least rank.
All elements of $x$ are in $D$ (because $D$ is transitive), and therefore they are also in $B$ (because $x$ has least rank among elements outside $B$). Therefore $x\subseteq B$ ...
Some sort of regularity or foundation is necessary, because there are non-well-founded counterexamples. For example, given an appropriate anti-foundation axiom, we can let $D=\{D\}$ and $B=\varnothing$. Then the premise is true, but $D\not\subseteq\varnothing$.