Under Foundation, $x \notin x$ 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: $x \notin x$ for every $x$, no sets satisfy $x \in y \in x$, and no sets satisfy $x \in y \in z \in x$.