Theorem 18.1 of Munkres’ Topology

Set $W= Y-V$. Then $W$ is closed in Y. Clearly $U= f^{-1}(W) = X- f^{-1}(V)$. To show that $U$ is closed, take a limit point $u$ of $U$. By the assumption of (2), $f(u)$ is in $W$. Thus $u$ must be in $U$. Hence $U$ is closed. Implying that $f^{-1}(V)$ is open.

$f: X\to Y$

(2) for every subset $A$ of $ X$, one has $f(\overline{A})\subseteq f(A) $

(3) for every closed set $B$ of $ Y$, the set $f^{-1}(B)$ is closed in $X$.

Proof :$ [(2) \implies (3)]$

Let, $B\subset Y$ is closed.

Suppose, $A=f^{-1}(B) $

Then by (2) , $f(\overline{f^{-1}(B)}\subseteq \overline{f(f^{-1}(B))}\subseteq \overline{B}=B$

Hence, $\overline{f^{-1}(B)}\subseteq {f^{-1}(B)}$

Note:

1. $C\subset D\implies \overline{C}\subseteq \overline{D}$

2. $f(f^{-1}(B)) \subseteq B $ for all $B\subseteq Y$

3. $C\subseteq \overline {C}$

Hence the conclusion follows immediately.

$(2) \implies (3)$ can be done as follows:

Suppose for all $A \subseteq X$ we have $f[\overline{A}]\subseteq \overline{f[A]}$. Let $B \subseteq Y$ be closed and define $A=f^{-1}[B]$.

Then apply the fact: $f[\overline{A}]\subseteq \overline{f[A]} \subseteq \overline{B} = B$ as $B$ is closed. So $\overline{A} \subseteq f^{-1}[B]=A$ and thus $A$ is closed and we've shown that the inverse image of the closed set $B$ is closed in $X$.

$(3) \iff (1)$ is trivial from $f^{-1}[Y\setminus A]= X\setminus f^{-1}[A]$ for all $A \subseteq Y$. If $A$ is open its complement is closed and vice versa. So inverse-preserving open sets is equivalent to inverse-preserving for closed sets and vice versa.

$(3) \implies (2)$ is also easy: $\overline{f[A]}$ is closed in $Y$ so $A \subseteq f^{-1}[f[A]]\subseteq f^{-1}[\overline{f[A]}]$ (trivial) has the right hand side closed so $\overline{A} \subseteq f^{-1}[\overline{f[A]}]$ and take $f$ on both sides (preserves the inclusion) to finish the proof.