Solution
= Solution
Because $f$ is <surjective>, $f^{-1}(B)$ is infinite whenever $B$ is infinite; a finite <preimage> could have only finitely many <image> points. By part (i), its <closure> is $X$. Apply part (a)(ii):
$$
Y=f(X)=f\bigl(\overline{f^{-1}(B)}\bigr)
\subseteq\overline{f(f^{-1}(B))}=\overline B.
$$
The other inclusion is automatic, so
$$
\boxed{\overline B=Y\quad\text{for every infinite }B\subseteq Y.}
$$
Both <continuity> and <surjectivity> enter the argument.