I get that using 2, 3, 5, 7 is sufficient to prove at most 23 primes in an interval of length 101, by straightforward but inelegant code:

  #!/usr/bin/env python3
  count = 101
  span = 101

  lp = [2, 3, 5, 7]

  for p1 in lp:
    count -= span//p1
    for p2 in lp:
        if p2 <= p1: continue
        count += span//(p1*p2)
        for p3 in lp:
            if p3 <= p2: continue
            count -= span//(p1*p2*p3)
            for p4 in lp:
                if p4 <= p3: continue
                count -= span//(p1*p2*p3*p4)
  print("At most {} primes in a span of {}".format(count,span))

A version of that which includes printouts of the number removed or added at each stage gives this output:

  Remove 14 divisible by 7
  Remove 20 divisible by 5
  Remove 33 divisible by 3
  Remove 50 divisible by 2

  Unremove 10 divisible by 2 * 5
  Unremove 16 divisible by 2 * 3
  Unremove 2 divisible by 5 * 7
  Unremove 4 divisible by 3 * 7
  Unremove 6 divisible by 3 * 5
  Unremove 7 divisible by 2 * 7

  Remove 0 divisible by 3 * 5 * 7
  Remove 1 divisible by 2 * 5 * 7
  Remove 2 divisible by 2 * 3 * 7
  Remove 3 divisible by 2 * 3 * 5

  Unremove 0 divisible by 2 * 3 * 5 * 7

  At most 23 primes in a span of 101

Hmm, good try, but it's not quite right. Here's a simpler way to get all the numbers divisible by 2, 3, 5, or 7, in a 100-number range, and which I think is easier to verify that it's correct:

    numbers_not_divisible_by_2_3_5_7 = []
    for i in range(520, 620):
        is_divisible = False
        for p in [2, 3, 5, 7]:
            if i % p == 0:
                is_divisible = True
                break
        if not is_divisible:
            numbers_not_divisible_by_2_3_5_7.append(i)

    print(numbers_not_divisible_by_2_3_5_7)
    print(len(numbers_not_divisible_by_2_3_5_7))

And if you run this code you'll find that it demonstrates a counterexample. Between 520 and 620 there are 25 numbers which are not divisible by 2, 3, 5, or 7:

[521, 523, 527, 529, 533, 541, 547, 551, 557, 559, 563, 569, 571, 577, 583, 587, 589, 593, 599, 601, 607, 611, 613, 617, 619]

You really need to use all the primes from 2 to 13 inclusive to get a limit of 23. And even then I was only able to prove it by brute-forcing up to the primorial of 13, but maybe there's a simpler way.

You are right. To fix mine, I'd need to take into account the actual endpoints of the interval instead of just the length of the interval. For each divisor d there would be a separate case to consider for each possible value of 10n mod d. This seems like it would greatly complicate my approach.