Let holes be numbered 1 thru 5. Inspecting the holes in any of the following sequences suffices:

2, 3, 4, 2, 3, 4
2, 3, 4, 4, 3, 2
4, 3, 2, 2, 3, 4
4, 3, 2, 4, 3, 2

Explanation for sequence 2, 3, 4, 2, 3, 4: Let F denote the set of holes where the rabbit might be hiding. On any morning, the rabbit is either in an even numbered hole or an odd numbered hole. So on the first morning, either F = {1, 3, 5} or F = {2, 4}. If F = {2, 4}, then the following sequence of inspections suffices to catch the rabbit: 2, 3, 4. However, if the rabbit was not caught, then F must have equalled {1, 3, 5} on the first morning, so F must equal {2, 4} on the fourth morning. Therefore, repeating the sequence 2, 3, 4 from the fourth day onwards would suffice to catch the rabbit.

Another explanation to convince us that the sequence 2, 3, 4, 2, 3, 4 suffices to catch the rabbit: Let F denote the set of holes where the rabbit might be hiding. Initially, F = {1, 2, 3, 4, 5}. If the rabbit is not in hole 2, it must move so that F = {2, 3, 4, 5}. If the rabbit is not in hole 3, it must move so that to F = {1, 3, 4, 5}. If the rabbit is not in hole 4, it must move so that F = {2, 4}. If the rabbit is not in hole 2, it must move so that F = {3, 5}. If the rabbit is not in hole 3, it must move so that F = {4}.