Does the set of open sets in a topological space have a topology itself?

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 compact-open 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 $X$:

1. $X$ is an exponentiable space in the category of topological spaces ($Y^X$ exists for all $Y$).


2. The exponential $Sigma^X$ exists.


3. The topology of $X$ is a continuous lattice.


4. The lattice of open sets of $X$ equipped with the Scott topology is the exponential $Sigma^X$.

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):

M.H. Escardo and R. Heckmann. Topologies on spaces of continuous functions. Topology Proceedings, volume 26, number 2, pp. 545-564, 2001-2002.