It may be better for you to consider uniform spaces instead of simply topological spaces. If you have a uniform space, then there is a very natural topology that one may put on the power set. Uniform spaces are closely related to topological spaces since one may go back and forth between topological and uniform spaces because uniform spaces are topological spaces with some extra structure. If $(X,\mathcal{U})$ is a uniform space, then $(X,\mathcal{U})$ induces a completely regular topology on $X$ where $U$ is open iff for each $x\in U$ there is an entourage $R\in\mathcal{U}$ with $R[x]\subseteq U$. Furthermore, every completely regular space $X$ can be given a compatible uniformity. For example, if $C$ is a compactification of $X$ such as the Stone-Cech compactification, then $C$ has a unique compatible uniformity. This uniformity on $C$ induces a uniformity on $X$.

Suppose that $(X,\mathcal{U})$ is a uniform space. Let $H(X)$ be the set of closed subsets of $X$. Then we can put a uniformity on $P(X)$ as follows. If $R\in\mathcal{U}$, then let $R^{\sharp}$ be the binary relation on $P(X)$ where $(A,B)\in R$ if and only if $A\subseteq R[B]=\{R[b]|b\in B\}$ and $B\subseteq R[A]$. Let $\widehat{R}$ be the restriction of $R^{\sharp}$ to $H(X)$. Then the system $\{R^{\sharp}|R\in\mathcal{U}\}$ generates a uniformity on $P(X)$, but this uniformity generally does not separate points of $X$. However, if we restrict this uniformity to $H(X)$, we get a separated uniformity on $H(X)$ and this uniformity is generated by the set of entourages $\{\widehat{R}|R\in\mathcal{U}\}$. This uniformity on $H(X)$ inherits some of the properties of your original uniform space $X$. For example, if $(X,d)$ is a metric space, then the hyperspace uniformity on $H(X)$ is induced by a metric $d^{\sharp}$ called the Hausdorff metric. The metric $d^{\sharp}$ is defined by \[d^{\sharp}(C,D)=\max[\sup_{c\in C}d(c,D),\sup_{d\in D}d(d,C)]\] \[=Max[\sup_{c\in C}\inf_{d\in D}d(c,d),\sup_{d\in D}\sup_{c\in C}d(c,d)].\] Furthermore, the Hausdorff metric $d^{\sharp}$ is complete whenever the original metric $d$ is complete. Now, if $C$ is a compact space, then $C$ can be given a unique uniform structure. With this uniform structure, the hyperspace $H(C)$ of $C$ remains compact. We say that a uniform space $(X,\mathcal{U})$ is non-Archimedean if it is generated by equivalence relations. The hyperspace of a non-Archimedean uniform space is always non-Archimedean.

The hyperspace uniformity is closely related to the Vietoris topology on a topological space which Steven Landsburg referred to in his answer. The reader is referred to Isbell's book on uniform spaces for more information about hyperspaces of uniform spaces.

It may be better for you to consider uniform spaces instead of simply topological spaces. If you have a uniform space, then there is a very natural topology that one may put on the power set. Uniform spaces are closely related to topological spaces since one may go back and forth between topological and uniform spaces because uniform spaces are topological spaces with some extra structure. If $(X,\mathcal{U})$ is a uniform space, then $(X,\mathcal{U})$ induces a completely regular topology on $X$ where $U$ is open iff for each $x\in U$ there is an entourage $R\in\mathcal{U}$ with $R[x]\subseteq U$. Furthermore, every completely regular space $X$ can be given a compatible uniformity. For example, if $C$ is a compactification of $X$ such as the Stone-Čech compactification, then $C$ has a unique compatible uniformity. This uniformity on $C$ induces a uniformity on $X$.

Suppose that $(X,\mathcal{U})$ is a uniform space. Let $H(X)$ be the set of closed subsets of $X$. Then we can put a uniformity on $P(X)$ as follows. If $R\in\mathcal{U}$, then let $R^{\sharp}$ be the binary relation on $P(X)$ where $(A,B)\in R$ if and only if $A\subseteq R[B]=\{R[b]|b\in B\}$ and $B\subseteq R[A]$. Let $\widehat{R}$ be the restriction of $R^{\sharp}$ to $H(X)$. Then the system $\{R^{\sharp}|R\in\mathcal{U}\}$ generates a uniformity on $P(X)$, but this uniformity generally does not separate points of $X$. However, if we restrict this uniformity to $H(X)$, we get a separated uniformity on $H(X)$ and this uniformity is generated by the set of entourages $\{\widehat{R}|R\in\mathcal{U}\}$. This uniformity on $H(X)$ inherits some of the properties of your original uniform space $X$. For example, if $(X,d)$ is a metric space, then the hyperspace uniformity on $H(X)$ is induced by a metric $d^{\sharp}$ called the Hausdorff metric. The metric $d^{\sharp}$ is defined by $$d^{\sharp}(C,D)=\max[\sup_{c\in C}d(c,D),\sup_{d\in D}d(d,C)]$$ $$=Max[\sup_{c\in C}\inf_{d\in D}d(c,d),\sup_{d\in D}\sup_{c\in C}d(c,d)].$$ Furthermore, the Hausdorff metric $d^{\sharp}$ is complete whenever the original metric $d$ is complete. Now, if $C$ is a compact space, then $C$ can be given a unique uniform structure. With this uniform structure, the hyperspace $H(C)$ of $C$ remains compact. We say that a uniform space $(X,\mathcal{U})$ is non-Archimedean if it is generated by equivalence relations. The hyperspace of a non-Archimedean uniform space is always non-Archimedean.

The hyperspace uniformity is closely related to the Vietoris topology on a topological space which Steven Landsburg referred to in his answer. The reader is referred to Isbell's book on uniform spaces for more information about hyperspaces of uniform spaces.