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If a set contains all its proper transitive subsets as members, do its members as well?

recently I came upon some personal notes I'd made several years ago while reviewing some basic set theory (ordinals, transfinite recursion, inaccessible cardinals etc.), and I stumbled upon a loose thread which I obviously had not resolved at the time, and which I would like to lay to rest:

Assuming some standard set theory (say ZF, even though I prefer NBG), without the Axiom of Foundation (preferably), one may define an ordinal $\alpha$ (von Neumann's definition) as a transitive set whose elements are well-ordered with respect to the membership relation $\in$. This is seen to be equivalent to the statement that $\alpha$ is transitive, all its $\beta\in\alpha$ are transitive too, and (as we cannot rely on foundation) for each non-empty $x\subseteq\alpha$ there exists some $\beta\in x$ such that $x\cap\beta=\emptyset$ (except for the last condition, this is as in Schofield's book on Mathematical Logic). One then goes on to prove that the class of all ordinals is well-ordered with respect to membership etc.; along the way a useful intermediate step is to prove that any ordinal $\alpha$ is (ad hoc definition) strange in the sense that one has $x\in\alpha$ for any transitive $x\subsetneq\alpha$.

My question finally (as this would provide an alternate definition of ordinal sets): are elements of strange sets themselves strange, or at least transitive ?

Thanks in advance for any useful comments ! Kind regards, Stephan F. Kroneck.