Topological Closure of Singleton is Irreducible/Proof 2

Proof
that
 * $\set x^-$ is not irreducible.

By Set is Subset of its Topological Closure:
 * $\set x \subseteq \set x^-$

By definitions of singleton and Subset:
 * $x \in \set x^-$

By definition of irreducible:
 * $\exists X_1, X_2 \subseteq S: X_1, X_2$ are closed

and:

By definition of union:
 * $x \in X_1$ or $x \in X_2$

By definitions of singleton and subset:
 * $\set x \subseteq X_1$ or $\set x \subseteq X_2$

By definition of closure :
 * $\set x^- \subseteq X_1$ or $\set x^- \subseteq X_2$

By Set is Subset of Union:
 * $X_1 \subseteq \set x$ and $X_2 \subseteq \set x$

Thus by definition of set equality:
 * this contradicts $\set x^- \ne X_1$ and $\set x^- \ne X_2$