Topological Closure of Singleton is Irreducible

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $x$ be a point of $T$.

Then $\left\{ {x}\right\}^-$ is irreducible

where $\left\{ {x}\right\}^-$ denotes the topological closure of $\left\{ {x}\right\}$.

Proof
Aiming for a contradiction suppose that
 * $\left\{ {x}\right\}^-$ is not irreducible.

By Set is Subset of its Topological Closure:
 * $\left\{ {x}\right\} \subseteq \left\{ {x}\right\}^-$

By definitions of singleton and Subset:
 * $x \in \left\{ {x}\right\}^-$

By definition of irreducible:
 * $\exists X_1, X_2 \subseteq S: X_1, X_2$ are closed $\land \left\{ {x}\right\}^- = X_1 \cup X_2 \land \left\{ {x}\right\}^- \ne X_1 \land \left\{ {x}\right\}^- \ne X_2$

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

By definitions of singleton and subset:
 * $\left\{ {x}\right\} \subseteq X_1$ or $\left\{ {x}\right\} \subseteq X_2$

By definition of Definition:Closure (Topology)/Definition 3:
 * $\left\{ {x}\right\}^- \subseteq X_1$ or $\left\{ {x}\right\}^- \subseteq X_2$

By Set is Subset of Union:
 * $X_1 \subseteq \left\{ {x}\right\}$ abd $X_2 \subseteq \left\{ {x}\right\}$

Thus by definition of set equality:
 * this contradicts $\left\{ {x}\right\}^- \ne X_1 \land \left\{ {x}\right\}^- \ne X_2$