Prove that $\mathcal T$ is a topology

We are first going to show that $h$ preserves inclusions. Suppose that $A\subset B$. This is equivalent to saying that there exists an $C\in \mathcal{P}(X)$ such that $A\cup C=B$. Applying $h$ to this equality, we find that $h(A\cup C)=h(A)\cup h(C)=h(B)$, here we used that $h$ goes through finite unions. Since $h(A)\cup h(C)=h(B)$, we have that $h(A)\subset h(B)$. Thus we showed that $A\subset B \Rightarrow h(A)\subset h(B)$. (The converse is not necessarily true).

Now choose $A_i^C\in \mathcal{T}$ arbitrarily. Then $\cap_i A_i\subset h(\cap_i A_i)$ by the third property of $h$. Notice that $\cap_i A_i\subset A_i$, applying $h$ to this inclusion, we get that $$h(\cap_iA_i)\subset h(A_i)=A_i$$ holds for all $i$. Here we used that $h(A_i)=A_i$ since $A_i^C\in \mathcal{T}$. Since the above property holds for all $i$, we conclude that $h(\cap_i A_i)\subset \cap_i A_i$. Thus $h(\cap_i A_i)=\cap_i A_i$, which we needed to show by my comment above. Notice that we didn't use that $h\circ h=h$.