To solve this problem, we need to consider the different cases in which the digits 2 and 5 do not come together in a 4-digit number formed using the digits 1, 2, 5, 6, 8, and 9 without repetition.

Case 1: The number starts with 2 or 5
If the number starts with 2, we have the following possibilities for the remaining three digits: 6, 8, and 9. This can be arranged in 3! = 6 ways.
If the number starts with 5, we have the following possibilities for the remaining three digits: 1, 6, 8, and 9. This can be arranged in 4! = 24 ways.

Case 2: The number does not start with 2 or 5
For the first digit, we have the following possibilities: 1, 6, 8, and 9. This can be arranged in 4 ways.
For the second digit, we have the following possibilities: 1, 6, 8, and 9 (excluding the digit used for the first digit). This can be arranged in 3 ways.
For the third digit, we have the following possibilities: 1, 6, 8, and 9 (excluding the digits used for the first two digits). This can be arranged in 2 ways.
For the fourth digit, we have the following possibilities: 1, 6, 8, and 9 (excluding the digits used for the first three digits). This can be arranged in 1 way.

Therefore, the total number of 4-digit numbers that can be formed without the digits 2 and 5 coming together is:

Case 1: 6 + 24 = 30
Case 2: 4 x 3 x 2 x 1 = 24

Total = 30 + 24 = 54

However, we need to consider that the question specifies that the digits cannot be repeated. Therefore, we need to divide this total by the number of ways the digits can be arranged, which is 4! = 24.

Final answer = 54 / 24 = 2.25

Since the number of 4-digit numbers must be a whole number, we round up to the nearest whole number, which is 3.

Therefore, the correct answer is 3 x 24 = 72, which corresponds to option C.