Suppose $f: \mathcal{P}(X) \to \mathcal{P}(X)$ is a function that
satisfies, for every set $A,B \subseteq X$
$(C_1): f(\emptyset) = \emptyset$
$(C_2): A \subseteq f(A)$
$(C_3): f(A \cup B) =f(A) \cup f(B)$
$(C_4): f(f(A)) = f(A)$
Prove that there exists a unique topology $\mathcal{T}$ on $X$ such that for all $A \subseteq X$, $\text{Cl}(A) = f(A)$, where $\text{Cl}(A)$ denotes the closure of the set $A$, with respect to the topology $\mathcal{T}$
My attempt:
I defined $\mathcal{T}:= \{A \subseteq X \mid f(X \setminus A) = X \setminus A\}$
and I managed to prove with $C_1, C_2, C_3$ that this set is a topology, and that it is unique, if it exists. So I only need to show that it exists, for which I have to prove that $\text{Cl}(A) = f(A)$ whenever $A \subseteq X$
I can show that $f(A) \supseteq \text{Cl}(A)$. I just need the other inclusion and I'm done!
Any hints?
I see that you define the closure of $A$ as $\operatorname{Cl}_{\mathcal{T}}(A) = \{x \in X: \forall O \in \mathcal{T}: x \in O \to O \cap A \neq \emptyset \}$, i.e. the set of adherent points of $A$.
For closure operators it's more convenient (and dual to the way we can define interiors) to use
$$\operatorname{Cl}_{\mathcal{T}}(A) = \bigcap \{F \subseteq X: A \subseteq F, F \text{ closed }\} (\ast)$$
which says that the closure of $A$ is the smallest closed set that contains $A$ as a subset.
It's a classic excercise in introductory topology to show that these definitions are equivalent.
Also the closure axioms (as in the case of interior operators) imply that $f$ is monotonic: $A \subseteq B$ implies $f(A) \subseteq f(B)$:
Proof: $A \subseteq B$ implies $A \cup B = B$ so $(C_3)$ tells us that $f(B) = f(A \cup B) = f(A) \cup f(B)$ so that $f(A) \subseteq f(B)$ follows.
Also note that in your definition we could just as easily have defined $A$ is closed iff $f(A) = A$.
Now if $A$ is a subset of $X$ we have that $A \subseteq f(A)$ by $(C_2)$ and $f(A)$ is closed by $(C_4)$, as $f(f(A)) = f(A)$. So $f(A)$ is one of the closed sets we take the intersection of in $(\ast)$ so that $\operatorname{Cl}_{\mathcal{T}}(A) \subseteq f(A)$.
On the other hand, if $F$ is one of the closed sets in the right hand side of $(\ast)$, we know that $A \subseteq F$ and $F$ is closed so that by monotonicity we know that $f(A) \subseteq f(F) = F$, and as this holds for all $F$, $f(A) \subseteq \operatorname{Cl}_{\mathcal{T}}(A)$ and we have equality of $f$ and the closure operator of $\mathcal{T}$ for any $A \subseteq X$.
Note the duality with your earlier interior operator here. Given an interior operator $i$ we can define a closure operator $c(A) = X\setminus i(X\setminus A)$. Standard set theory gives that axioms $(I_1)-(I_4)$ there then are equivalent to $(C_1)-(C_4)$ above. We can reduce one excercise to the other, essentially. $i(A) = X\setminus c(X\setminus A)$ allows us to go from a closure operator to an interior operator in the same way.