 Solution

Assume a nonempty set $a$ has no membership-minimal element, so for every $x\in a$ there is $y\in x\cap a$. Consider the formula $\phi (x)\equiv x\notin a$.

It is progressive: if every element of $x$ is outside $a$ but $x\in a$, the assumed absence of a minimal element supplies $y\in x\cap a$, contradicting that hypothesis. The membership induction theorem therefore implies $x\notin a$ for every set $x$. This contradicts $a\ne \varnothing$.

Thus every nonempty set $a$ contains an element $x$ with $x\cap a=\varnothing$, exactly the Axiom of foundation:Only the induction schema and elementary set reasoning were needed; no prior use of Axiom of foundation occurs in the proof.