Solution
= Solution
Any proper <closed subset> $F$ of $Y$ must be finite: if it were infinite, part (ii) would give $F=\overline F=Y$. Therefore every nonempty <open subset> of $Y$, being the <complement> of a proper <closed set>, is cofinite.
$$
\boxed{\mathcal T_Y\subseteq\mathcal T_{\mathrm{cofinite}}.}
$$
The empty set belongs to both topologies. Inclusion need not be equality; the target may have fewer <open> sets.