How to prove that if $A\subseteq A\times A$, then $A=\varnothing$?

How do I prove that $A\subseteq (A \times A)\rightarrow A = \varnothing$ in axiomatic set theory?

This is theorem 107 of Patrick Suppes "Axiomatic Set Theory". I do not buy his argument that the set $A$ does not contain any empty sets because, according to him, the set $A$ is a subset of $A\times A$. Are there any other arguments that prove this theorem? Why does the set $A$ not have the empty set as a member?

Suppose towards a contradiction that $A\subseteq A\times A$ and $A$ is nonempty. Pick an element $a$ of $A$ of minimal rank. Then $a=(b, c)$ for some $b, c\in A$, since every element of $A$ is an ordered pair of elements of $A$ (this is just "$A\subseteq A\times A$" in words).

But then $b$ and $c$ each have rank strictly less than that of $a$, which is a contradiction.

Note that this depends on the precise definition of the pairing function; for some pairing functions we may in fact have $A\subseteq A\times A$ for nonempty $A$. In particular, maybe we choose to represent the pair $\langle 17, 17\rangle$ by $17$; then $A=\{17\}$ satisfies $A=A\times A$. The result Suppes states is wihtin the context of one of the usual ordered pair notions.