Exists Element Not in Set/Proof 2

Proof
By Axiom of Specification, we can construct the set:
 * $T = \set {x \in S: x \notin x}$

Then for all $y$:
 * $(*) \quad y \in T$ $\paren {y \in S \land y \notin y}$.

$T \in S$.

By Law of Excluded Middle, either $T \in T$ or $T \notin T$.

Suppose $T \in T$.

By $(*)$, $T \in S \land T \notin T$.

By Rule of Simplification we have $T \notin T$, which is a contradiction.

Now suppose $T \notin T$.

By $(*)$ again, we have $\neg \paren {T \in S \land T \notin T}$.

By Modus Ponendo Tollens, $\neg \paren {T \in S}$, which is also a contradiction.

Hence we must have $T \notin S$.

And thus there exists something (namely $T$) that does not belong to $S$.