Proof of No Set Is an Element of Itself, Nor Is Its Singleton, and No Set Equals Its Singleton

Proof of (i). The singleton $\{X\}$ is a set whose only element is $X$ (The Empty Set, the Unordered Pair and the Singleton (iii)). By Axiom of Foundation for Classes, applied to $\{X\}$, which has the element $X$, some element $u$ of $\{X\}$ is such that no set lies in both $u$ and $\{X\}$; necessarily $u=X$. If $X\in X$, then the set $X$ would lie in both $X$ and $\{X\}$, which is impossible. Hence $X\notin X$.

Proof of (ii). Let $P=\{X,\{X\}\}$ be the unordered pair of the sets $X$ and $\{X\}$, whose elements are exactly $X$ and $\{X\}$. By Axiom of Foundation for Classes, applied to $P$, some $u\in P$ is such that no set lies in both $u$ and $P$. If $u=\{X\}$, then the set $X$ lies in $u$ and in $P$, which is impossible; so $u=X$. As $\{X\}\in P$, it follows that $\{X\}\notin X$.

Proof of (iii). $X\in\{X\}$, while $X\notin X$ by (i); so $X$ and $\{X\}$ do not have the same elements, and $X\neq\{X\}$.