Under Foundation, $x\mathrm{\notin }x$ for every set $x$, there are no sets with $x\in y\in x$, and there are no sets with $x\in y\in z\in x$

Suppose the statement fails, so that at least one of the following holds: there is a set $x$ with $x\in x$; there are sets $x$ and $y$ with $x\in y$ and $y\in x$; there are sets $x$, $y$ and $z$ with $x\in y$, $y\in z$ and $z\in x$.

Suppose $x\in x$ and put $S:=\{x\}$. Its only member is $x$, so $S$ has a member and Foundation applies; the member it supplies must be $x$, so $x$ shares no member with $S$. But $x\in x$ and $x\in S$, so $x$ is a member of both.

Suppose $x\in y$ and $y\in x$, and put $S:=\{x,y\}$. Foundation supplies a member of $S$ sharing no member with $S$, and that member is $x$ or $y$. If it is $x$, then $y\in x$ and $y\in S$ make $y$ a member of both; if it is $y$, then $x\in y$ and $x\in S$ make $x$ a member of both.

Suppose $x\in y$, $y\in z$ and $z\in x$, and put $S:=\{x,y\}\cup \{z\}$, whose members are exactly $x$, $y$ and $z$. Foundation supplies a member of $S$ sharing no member with $S$. If it is $x$, then $z\in x$ and $z\in S$; if it is $y$, then $x\in y$ and $x\in S$; if it is $z$, then $y\in z$ and $y\in S$. In each case that member shares a member with $S$.

Each of the three suppositions contradicts the member Foundation supplies, so none of them holds: $x\mathrm{\notin }x$ for every $x$, no sets satisfy $x\in y\in x$, and no sets satisfy $x\in y\in z\in x$.