Theorem 18.1 of Munkres’ Topology 
Let $X$ and $Y$ be topological spaces; let $f:X\to Y$. Then the following are equivalent:
(1) $f$ is continuous
(2) for every subset $A$ of $X$, one has $f(\overline{A}) \subseteq \overline{f(A)}$.
(3) for every closed set $B$ of $Y$, the set $f^{-1}(B)$ is closed in $X$.
(4) for each $x\in X$ and every neighbourhood $V$ of $f(x)$, there is a neighbourhood $U$ of $x$ such that $f(U)\subseteq V$.

Now there are lots of way to prove this theorem, like $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (4) \Rightarrow (1)$, I think this is the least number of implication needed to prove this theorem(Note I’m not saying it is the most easiest way to do it) and $(1)\Leftrightarrow (2)$, $(1) \Leftrightarrow (3)$, $(1)\Leftrightarrow (4)$, this is maybe the most number of implication needed to prove this theorem. You would agree proof of this $(1)\Leftrightarrow (3)$ is trivial/easy and $(1)\Leftrightarrow (4)$ is decent exercise. We have already show $(1)\Rightarrow (2)$. How do we show $(2)\Rightarrow (1)$?
Can we use the similar approach as in the proof of $(2)\Rightarrow (3)$? i.e. let $V\in \mathcal{T}_{Y}$. We need to show $X-f^{-1} (V)$ is closed in $X$. Put $A= X-f^{-1} (V)$. Let $x\in \overline{A}$. So $f(x)\in f(\overline{A})\subseteq \overline{f(A)}$. $f(A) \supseteq f(X) - V= Y- V$. Thus $Y-V \subseteq \overline{f(A)}
$. Here I’m stuck.
 A: 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.
A: $f: X\to Y$
(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:

*

*$C\subset D\implies \overline{C}\subseteq  \overline{D}$


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


*$C\subseteq \overline {C}$
Hence the conclusion follows immediately.
A: $(2) \implies (3)$ 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.
