Given, we need to find the number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once.

Approach:
We can form the integers in the following way:

- For integers between 2000 and 2999: We can select any of the 6 digits for the thousands place (since 0 cannot be the first digit), and any of the remaining 5 digits for the hundreds place, any of the remaining 4 digits for the tens place, and any of the remaining 3 digits for the units place. Thus, we have 6 × 5 × 4 × 3 = 360 integers in this range.
- For integers between 3000 and 4999: We can select any of the 4 digits other than 0 for the thousands place, any of the remaining 5 digits for the hundreds place, any of the remaining 4 digits for the tens place, and any of the remaining 3 digits for the units place. Thus, we have 4 × 5 × 4 × 3 = 240 integers in this range.
- For integers between 5000 and 5999: We can select any of the 2 digits other than 0 for the thousands place, any of the remaining 5 digits for the hundreds place, any of the remaining 4 digits for the tens place, and any of the remaining 3 digits for the units place. Thus, we have 2 × 5 × 4 × 3 = 120 integers in this range.

Therefore, the total number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once is 360 + 240 + 120 = 720 + 720 = 1440.

Hence, the correct answer is option B, 1440.

To solve this problem, we need to consider the different possible cases and count the number of integers that can be formed.

Case 1: The thousands digit is 2, 3, 4, or 5:
- In this case, the thousands digit can be chosen in 4 ways.
- The hundreds digit can be chosen in 6 ways (any of the remaining 6 digits).
- The tens digit can be chosen in 5 ways (any of the remaining 5 digits).
- The units digit can be chosen in 4 ways (any of the remaining 4 digits).
- So, the number of integers in this case is 4 x 6 x 5 x 4 = 480.

Case 2: The thousands digit is 1:
- In this case, the thousands digit can be chosen in 1 way (as 1).
- The hundreds digit can be chosen in 6 ways (any of the remaining 6 digits).
- The tens digit can be chosen in 5 ways (any of the remaining 5 digits).
- The units digit can be chosen in 4 ways (any of the remaining 4 digits).
- So, the number of integers in this case is 1 x 6 x 5 x 4 = 120.

Case 3: The thousands digit is 0:
- In this case, the thousands digit can be chosen in 1 way (as 0).
- The hundreds digit can be chosen in 6 ways (any of the remaining 6 digits).
- The tens digit can be chosen in 5 ways (any of the remaining 5 digits).
- The units digit can be chosen in 4 ways (any of the remaining 4 digits).
- So, the number of integers in this case is 1 x 6 x 5 x 4 = 120.

Total number of integers:
- To find the total number of integers, we need to sum up the number of integers from each case.
- Total number of integers = 480 + 120 + 120 = 720.

However, we need to consider that the problem asks for integers greater than 2000. So, we need to subtract the number of integers that are less than or equal to 2000.

Number of integers less than or equal to 2000:
- The thousands digit can be chosen in 2 ways (0 or 1).
- The hundreds digit can be chosen in 6 ways (any of the remaining 6 digits).
- The tens digit can be chosen in 5 ways (any of the remaining 5 digits).
- The units digit can be chosen in 4 ways (any of the remaining 4 digits).
- So, the number of integers less than or equal to 2000 is 2 x 6 x 5 x 4 = 240.

Final answer:
- The number of integers greater than 2000 that can be formed is 720 - 240 = 480.
- Therefore, the correct answer is option B) 1440.