G being finite.
Since you have proven that the intersection of H and K is a group, you are done! From the proof of Lagrange, this subgroup has a particular (finite) index.
G being infinite. A good example of this that I can think of is G = Z with the + operation and subgroups H,K being the {3n | n in Z} and {4n | n in Z}. These have finite indexes in G. (Eg, the left cosets of H are {3n | n in Z}, {3n+1 | n in Z}, and {3n+2 | n in Z}.) The intersection of H and K is {12n | n in Z} which has index 12 in G - clearly finite.
The only problem is that it would seem difficult to generalise this example to other infinite groups. Here I would turn to contradiction.
Suppose that H and K have finite indexes in G but their intersection has an infinite index. The intersection of H and K is a subgroup of H. At this point, I'm going to point out that K is irrelevant - K is only here to give us a subset of H (intersection of H,K), which is actually a group (so that we can talk about the index of this subset). If a subset (subgroup) of H has infinite index... How can H possibly have a finite index?
Posted from TSR Mobile