Closure of Irreducible Subspace is Irreducible/Proof 2

Proof
In view of, it suffices to show that for all closed sets $A_1$ and $A_2$ in $T$:
 * $Y^- \subseteq A_1 \cup A_2 \implies \exists i_0 \in \set {1, 2} : Y^- \subseteq A_{i_0}$

To this end, let $A_1$ and $A_2$ be closed sets in $T$ such that:
 * $Y^- \subseteq A_1 \cup A_2$

Then, in particular:
 * $Y \subseteq A_1 \cup A_2$

Since $Y \subseteq S$ is an irreducible subset:
 * $\exists i_0 \in \set {1, 2} : Y \subseteq A_{i_0}$

By Closure of Subset of Closed Set of Topological Space is Subset:
 * $Y^- \subseteq A_{i_0}$