Suppose we have a co-topology $\kappa$ and we define
$$c(A) = \bigcap \{C \in \kappa: A \subseteq C\}$$
so that $c(A) \in \kappa$ ($\kappa$ is closed under intersections) and $A \subseteq c(A)$. The intersection is non-void because $X \in \kappa$ is always one of the sets in the intersection.
We want to see that $c(A \cup B) = c(A) \cup c(B)$.
It's already clear from the definition that $A \subseteq B$ implies $c(A) \subseteq c(B)$. (the sets from $\kappa$ that contain $B$ also contain $A$, so $c(A)$ is the intersection of possibly more sets than $c(B)$ hence smaller).
So $A \subseteq c(A), B \subseteq c(B)$ and so $A \cup B \subseteq c(A) \cup c(B)$. The right hand side is in $\kappa$ as this is closed under finite unions, and as it contains $A \cup B$, $c(A) \cup c(B)$ is one of the sets being intersected in the definition of $c(A \cup B)$ and so
$$c(A \cup B) \subseteq c(A) \cup c(B)$$
On the other hand $A,B \subseteq A \cup B$ so $c(A), c(B) \subseteq c(A \cup B)$
and it follows that
$$c(A) \cup c(B) \subseteq c(A \cup B)$$
Hence we have equality.
We also have that $c(A) = A$ iff $A \in \kappa$, to answer the final question affirmatively. $c(A) = A$ implies $A \in \kappa$ because $c(A) \in \kappa$ always. And if $A \in \kappa$, $A$ itself is in the intersecting family defining $c(A)$ so that $A \subseteq c(A) \subseteq A$ and $c(A) = A$.
Your definition of $\tau$ isn't right. By your definition $\tau$ is simply the powerset of $X$. Here is likely what you want:
$$\tau=\{X\setminus h(A)\mid A\subseteq X\}$$
Then $X\setminus h(\emptyset)=X\in\tau$ and $X\setminus h(X)=\emptyset\in\tau$.
If $A,B\in\tau$ then $A=X\setminus h(U)$ and $B=X\setminus h(V)$ for some $U,V\subseteq X$. Then
$$A\cap B=(X\setminus h(U))\cap(X\setminus h(V))=X\setminus(h(U)\cup h(V))=X\setminus h(U\cup V)$$
Therefore $A\cap B\in\tau$.
Finally if $\{A_{\alpha}\}_{\alpha\in I}$ is some collection of elements in $\tau$ then say that $A_{\alpha}=X\setminus h(U_{\alpha})$ for some $U_{\alpha}\subseteq X$. Then
$$\bigcup_{\alpha\in I}A_{\alpha}=\bigcup_{\alpha\in I}(X\setminus h(U_{\alpha}))=X\setminus\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)$$
To finish this line we will need your third and second axioms.
$$\bigcap h(U_{\alpha})\subseteq h\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)$$
For each $\alpha\in I$ we have that $\bigcap_{\beta\in I}h(U_{\beta})\subseteq h(U_{\alpha})$
Thus by monotonicity and idempotence (axioms two and three) we have
$$h\left(\bigcap_{\beta\in I}h(U_{\beta})\right)\subseteq hh(U_{\alpha})=h(U_{\alpha})$$
for each $\alpha$. We then have that
$$h\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)\subseteq\bigcap_{\alpha\in I}h(U_{\alpha})\subseteq h\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)$$
Establishing equality between the two sets. We then have that
$$X\setminus\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)=X\setminus h\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)$$
Therefore, the union over the family $\{A_{\alpha}\}_{\alpha\in I}$ is an element of $\tau$, establishing that $\tau$ is indeed a topology on $X$.
Best Answer
The definition of the topology $\tau$ is fine, now you have to show first that the closure $\operatorname{cl}_\tau$ of taking the closure in $\tau$ actually equals $f$, so that for all subsets of $X$ we have $\operatorname{cl}_\tau(A)= f(A)$.
So fix $A$. First note that a set is closed in $\tau$ iff it is of the form $f(B)$ for some $B$, and $\operatorname{cl}_\tau(A)$ is defined as $$\operatorname{cl}_\tau(A) = \bigcap \{C: C \text{ closed in } \tau, A \subseteq C\}$$
the smallest $\tau$-closed containing $A$.
Now $f(A)$ is closed by definition of $\tau$ and one axiom of the closure tellsus that $A \subseteq f(A)$, so $f(A)$ is one of the $\tau$-closed sets in the intersection that defines $\operatorname{cl}_\tau(A)$ so from that it follows that
$$\operatorname{cl}_\tau(A) \subseteq f(A)$$
as the intersection is a subset of each of the sets it's an intersection of, by definition.
On the other hand if $C$ is any closed set containing $A$, we know $A \subseteq C$ and $C=f(B)$ for some $B$, as said. But then monotonicity of $f$ (e.g. follows from union axiom), and idempotence of $f$ tells us that $f(A) \subseteq f(C)= f(f(B))=f(B)=C$, and as $C$ was arbitrary, $f(A)$ is a subset of all sets in the definining intersection of $\operatorname{cl}_\tau(A)$ so
$$f(A) \subseteq \operatorname{cl}_\tau(A)$$ and we have the equality, we wanted.
If $\tau'$ is a topology on $X$ so that $f(A)=\operatorname{cl}_{\tau'}(A)$ for all $A$, then we have that $\operatorname{cl}_\tau(A)=\operatorname{cl}_{\tau'}(A)$ for all $A$ for the $\tau$ we defined. And so $\tau'$ has the same closed sets as $\tau$ (a set is closed iff it equals its closure) and so the same open sets too (as a set is open iff its complement is closed). So $\tau=\tau'$ and we have unicity.
The unicity argument alone doesn't need us to define $\tau$, it follows quite easily by the above argument that topologies that induce the same closure operation for all sets, are the same topologies.
In formulae, if the topologies $\tau,\tau'$ both "work" for the given $f$:
$$O \in \tau \iff O^\complement \text{ closed in }\tau \iff O^\complement=\operatorname{cl}_\tau(O^\complement)=f(O^\complement)=\operatorname{cl}_{\tau'}(O^\complement) \iff \\ O^\complement \text{ closed in }\tau' \iff O \in \tau'$$