How can I golf this code to count all perfect squares below 100?

For the exact question posed: since we know that 1 is a perfect square and all integers between that and the maximal one (here 9) will be included we can simply find that maximal one:

print((100-1)**.5//1)

(//1 performing integer division by one to remove any fractional part may be replaced with /1 prior to Python 3.)

with both endpoints (an inclusive start and exclusive stop equivalent to a range) this could be extended to a function (negative inputs catered for with max):

f=lambda start, stop:print(max(0,stop-1)**.5//1-max(0,start)**.5//1)

For the record, below is another approach using additions and multiplications only.

The square of N is the sum of the N first odd positive integers:

1^2 = 1
2^2 = 1 + 3 = 4
3^2 = 1 + 3 + 5 = 9
4^2 = 1 + 3 + 5 + 7 = 16
etc.

Consequently, if we are to compute all perfect squares up to a given limit, each one can be quickly deduced from the previous one.

Hence the following possible algorithms:

# with 3 variables, using addition only
s = i = 1
n = 0

while s < 100:
  n += 1
  i += 2
  s += i

print(n)

# with 2 variables, using addition and multiplication
s = 1
n = 0

while s < 100:
  n += 1
  s += n * 2 + 1

print(n)

Or as a recursive lambda:

f = lambda x, s=0, n=0: f(x, s+n*2+1, n+1) if s < x else n-1

print(f(100))