Closure of Irreducible Subspace is Irreducible/Proof 2

Proof
Observe that for each subset $V \subseteq Y^-$ which is closed in $T$:
 * $(1): \quad Y^- \setminus V \ne \O \implies Y \setminus V \ne \O$

Indeed:

$Y^-$ is not irreducible.

That is, there exist proper subsets $V_1, V_2$ of $Y^-$ which are closed subsets in $T$ such that:
 * $Y^- = V_1 \cup V_2$

Then, from $(1)$:
 * $Y \setminus V_1 \ne \O$

and:
 * $Y \setminus V_2 \ne \O$

In particular:
 * $V_1 \cap Y \subsetneqq Y$

and:
 * $V_2 \cap Y \subsetneqq Y$

Because:
 * $Y = \paren {V_1 \cap Y} \cup \paren {V_2 \cap Y}$

we have also:
 * $V_1 \cap Y \ne \O$

and:
 * $V_2 \cap Y \ne \O$

Therefore $V_1 \cap Y$ and $V_2 \cap Y$ are proper closed subsets of $Y$, that form a cover of $Y$.

This contradicts the fact that $Y$ is irreducible.