Closure of Irreducible Subspace is Irreducible/Proof 3

Proof
Observe that for each closed set $V$ in $T$:
 * $(1): \quad V \subsetneqq Y^- \implies V \cap Y \subsetneqq Y$

Indeed:

$Y^-$ is not irreducible.

That is, there exist $V_1, V_2 \subsetneqq Y^-$, closed in $\struct {Y^-, \tau_{Y^-} }$, such that:
 * $Y^- = V_1 \cup V_2$

$V_1$ and $V_2$ are also closed in $T$, since $Y^-$ is closed in $T$,

Then, from $(1)$:
 * $V_1 \cap Y \subsetneqq Y$

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

Therefore $V_1 \cap Y$ and $V_2 \cap Y$ are proper subsets of $Y$ such that:
 * $Y = \paren {V_1 \cap Y} \cup \paren {V_2 \cap Y}$

Furtheremore, $V_1 \cap Y$ and $V_2 \cap Y$ are closed in $\struct {Y, \tau_Y}$.

This contradicts the fact that $Y$ is irreducible.