How to solve the question in the following picture. Here, the Tychonoff topology on the power set $\mathcal{P}(X)$ of a set $X$ has as its subbasis sets in $\mathcal{P}(\mathcal{P}(X))$ of the form $U_{x}=\{A \subset X : x \in A\}$ and $U_{x}^{c}=\{A \subset X: x \notin A\}$. So a basis for it is sets that are finite intersections of those two types of sets mentioned above. So I need to show that $\mathcal{P}(A)^{c}$ is arbitrary unions of finite intersections of those two types of sets mentioned above.
For a fixed $A \subset X$, there is a natural inclusion $\mathcal{P}(A) \subset \mathcal{P}(X)$. Show that $\mathcal{P}(A)$ is a closed subset.
The inclusion $i$ is obvious as a map: if $B \in \mathcal{P}(A)$, then $B \subseteq A \subseteq X$ so $i(B) = B \in \mathcal{P}(X)$.
This map $i: \mathcal{P}(A) \to \mathcal{P}(X)$ is continuous, because for any $x \in X$: $i^{-1}[U_x] =\{B \subseteq A: i(B) \in U_x\} = \{B \subseteq A: x \in B\} = U_x(A)$, where $U_x(A)$ is the subbasic set as you defined, but for $\mathcal{P}(A)$ instead of $\mathcal{P}(X)$.
A similar fact holds for $U_x^c$ as well, so $i$ has the property that inverse images of subbasic sets are open, so $i$ is continuous.
To see that $\mathcal{P}(A)$ (or really its image under $i$) is closed in $\mathcal{P}(X)$: suppose $C \notin \mathcal{P}(A)$. This means there is a point $p \in C$ such that $p \notin A$. But then the subbasic set $U_p$ of $\mathcal{P}(X)$ also has the property that all its members by definition contain this "bad" point $p$, so are not subsets of $A$. So $C \in U_x \subseteq \mathcal{P}(X)\setminus \mathcal{P}(A)$, so that the complement of $\mathcal{P}(A)$ is open.