If X is a topological space, and A consists of all of X's open sets, can we define a natural topology on A (using the topology of X)?

Of course there are many answers to your question. The interesting thing to ask is if there is a "best" or "right" answer. In many respects the "correct" topology for the lattice of open sets is the Scott topology. In case $X$ is locally compact, the Scott topology coincides with the compactopen topology of the continuous function space $C(X,\Sigma)$, where $\Sigma$ is the Sierpinski space (where we identify open sets with their characteristic functions into $\Sigma$). There are several reasons why the Scott topology is the "right" one. One of them is that the following are equivalent for a space $X$:
I recommend the following paper by Martin Escardó and Reinhold Heckmann in which they explain many things related to topology of the lattice of open sets (and function spaces in general):



If $X$ is compact Hausdorff then the Vietoris topology (Wikipedia is lacking here, consult your standard topology textbook) on the compact (i.e. closed) subsets of $X$ implicitly defines a compact Hausdorff topology on the open subsets of $X$ via complements. 


The topology is a preorder/post/lattice (amongst other things), and there are various topologies one can put on lattices: the Scott topology the Lawson topology In general domain theory brings up lots of things along this line 


If $X$ is a metric space, you can use Hausdorff distance to get a metric on the closed sets. 

