Number of subsets might be exponential (max 2^k), so there is nothing wrong to traverse all possible independent subsets with recursion. I've used linear search of the next possible interval, but it is worth to exploit binary search.
def nonovl(l, idx, right, ll):
if idx == len(l):
if ll:
print(ll)
return
#find next non-overlapping interval without using l[idx]
next = idx + 1
while next < len(l) and right >= l[next][0]:
next += 1
nonovl(l, next, right, ll)
#find next non-overlapping interval after using l[idx]
next = idx + 1
right = l[idx][1]
while next < len(l) and right >= l[next][0]:
next += 1
nonovl(l, next, right, ll + str(l[idx]))
l=[(1,8),(2,13),(9,18),(15,30),(20,35)]
l.sort()
nonovl(l, 0, -1, "")
(20, 35)
(15, 30)
(9, 18)
(9, 18)(20, 35)
(2, 13)
(2, 13)(20, 35)
(2, 13)(15, 30)
(1, 8)
(1, 8)(20, 35)
(1, 8)(15, 30)
(1, 8)(9, 18)
(1, 8)(9, 18)(20, 35)
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