Closure of Irreducible Subspace is Irreducible/Proof 2

Proof
Observe that for each closed subset $A \subseteq Y^-$:
 * $Y^- \setminus A \ne \O \implies Y \setminus A \ne \O$

Indeed:

$Y^-$ is not irreducible.

That is, there exist proper closed subsets $A,B$ of $Y^-$ such that:
 * $Y^- = A \cup B$

Then, from the above observation:
 * $Y \setminus A \ne \O$

and:
 * $Y \setminus B \ne \O$

In particular:
 * $A \cap Y \subset Y$

and:
 * $B \cap Y \subset Y$

Since:
 * $Y = \paren {A \cap Y} \cup \paren {B \cap Y}$

we have also:
 * $A \cap Y \ne \O$

and:
 * $B \cap Y \ne \O$

Therefore $A \cap Y$ and $B \cap Y$ are proper closed subsets of $Y$, that form a cover of $Y$.

This contradicts the fact that $Y$ is irreducible.