1-7 together are pretty strong. You aren't going to have a procedure for doing this without most of the powerset topologies being indiscrete.
First note that if $X$ is a set and $\tau$ is at topology on $\mathcal{P}(X)$ satisfying 2-4, then the function $A \mathbin{\Delta} B = (A \cup B) \setminus (A \cap B)$ is continuous, which implies that for fixed $B\subseteq X$, the map $A \mapsto A \mathbin{\Delta} B$ is a homeomorphism. Therefore for any $A$ and $B$ in $\mathcal{P}(X)$, there is a homeomorphism of $(\mathcal{P}(X),\tau)$ taking $A$ to $B$ (namely the map $C \mapsto C \mathbin{\Delta} (A \mathbin{\Delta} B)$).
When I say a topology is non-trivial, I mean specifically that it is not indiscrete.
Lemma. If $X$ is a set and $(\mathcal{P}(X),\tau)$ satisfies 2-4, then the closure $F$ of $\{\varnothing\}$ is a filter (i.e., $A \subseteq B \in F \Rightarrow A \in F$ and $A\in F \wedge B \in F \Rightarrow A \cup B \in F$). In particular, $\tau$ is non-trivial if and only if the $X \notin F$.
Proof. Suppose that $A \notin F$ and $B \supseteq A$. Since $\cap$ is continuous, we have that $G = \{C : C \cap A \in F\}$ is closed. Clearly $\varnothing \in G$, so $F \subseteq G$. Furthermore, $B \notin G$, so $B \notin F$.
For showing that $F$ is closed under unions, first note that it is sufficient to show it for disjoint $A,B \in F$. So assume that $A,B \in F$ and $A$ and $B$ are disjoint. Since $C \mapsto C \mathbin{\Delta} B$ is a homeomorphism, we have that $A \in F$ if and only if $A \mathbin{\Delta} B \in \overline{\{B\}}$ (since $\varnothing \mathbin{\Delta} B = B$). Therefore $A \mathbin{\Delta} B \in \overline{\{B\}} \subseteq F$, as required.
For the final statement. Since $\mathcal{P}(X)$ has a transitive homeomorphism group, $\tau$ is non-trivial if and only if $F$ is not all of $\mathcal{P}(X)$. Since $F$ is a filter, this happens if and only if $X \notin F$. $\square_{\text{Lemma}}$
Proposition. For any topological space $X$, if $(\mathcal{P}(X),\tau_1)$ and $(\mathcal{P}(\mathcal{P}(X)),\tau_2)$ satisfy 1-5 and 7 and $\tau_1$ is non-trivial, then $X$ is discrete.
Proof. To hopefully make this proof a little bit easier to read, I'm going to denote the empty set as $\varnothing_1$ when we're thinking about it as an element of $\mathcal{P}(X)$ and as $\varnothing_2$ when we're thinking about it as an element of $\mathcal{P}(\mathcal{P}(X))$.
Let $F_1$ be the $\tau_1$-closure of $\{\varnothing_1\}$, and let $F_2$ be the $\tau_2$-closure of $\{\varnothing_2\}$. Since $\tau_1$ is non-trivial, $X \notin F_1$ by the lemma. Since $\bigcup: \mathcal{P}(\mathcal{P}(X)) \to \mathcal{P}(X)$ is continuous, we have that $\bigcup^{-1}(F_1)$ is a closed set in $\mathcal{P}(\mathcal{P}(X))$. Since $\varnothing_2 \in \bigcup^{-1}(F_1)$, we have that $F_2 \subseteq \bigcup^{-1}(F_1)$, i.e., if $A \in F_2$, then $\bigcup A \in F_1$. Therefore, in particular, $\{X\} \notin F_2$.
By 7, every homeomorphism $f$ of $\mathcal{P}(X)$ induces a homeomorphism $f_\ast$ of $\mathcal{P}(\mathcal{P}(X))$. Each such homeomorphism $f_\ast$ fixes $\varnothing_2$ and takes singletons to singletons (specifically, for any $A \in \mathcal{P}(X)$, $f_\ast(\{A\}) = \{f(A)\}$). By the discussion before the lemma, the homeomorphism group of $\mathcal{P}(X)$ is transitive. Therefore $\{A\} \notin F_2$ for every $A \in \mathcal{P}(X)$ and so $F_2 = \{\varnothing_2\}$ (because $F_2$ does not contain one particular singleton, $\{X\}$, so it does not contain any singletons, since it needs to be invariant under any homeomorphisms of $\mathcal{P}(\mathcal{P}(X))$ fixing $\varnothing_2$).
Consider the function $(x,y) \mapsto \{x\} \cap \{y\}$ from $\mathcal{P}(X) \times \mathcal{P}(X) \to \mathcal{P}(\mathcal{P}(X))$. By 1 and 3, this is continuous. Since $\{\varnothing_2\}$ is closed, this implies that the off diagonal $\{(x,y) \in \mathcal{P}(X) \times \mathcal{P}(X): x \neq y\}$ is closed and therefore the diagonal is open. It's relatively straightforward to show that if the diagonal of $Y^2$ is open, then $Y$ has the discrete topology. Therefore $\tau_1$ is discrete. Repeating the argument again (since now we know that $\{\varnothing_1\}$ is closed) gives that the topology on $X$ is discrete. $\square$
