prove that in any perfect square the digits immediately to the left of units digit cannot be 101

This makes use of the fact that every square is of the form 4k or 4k+1.

Remember that the digit in the units place of a perfect square would belong to {0,1,4,5,9}

If the last digit is 0, then the last 4 digits are 1010 and hence the remainder on division by 4 is 2.

In this way we see that the remainder set is {2,3}. Hence no perfect squares exist such that number 101 appears before the units digit

you have considered only four digit numbers but the question is general