I am going to prove the stronger:
When you put different whole numbers on a $7 \times 7$ board there has to be a pair of adjacent squares with a difference of at least $5$.
Proof by contradiction: Suppose there is no such adjacent pair of squares, i.e all adjacent squares have a difference of at most $4$.
Let $n$ be the number in the center square.
Given that adjacent squares have a difference of at most $4$, that means that the numbers in the squares of the sub-board can at most differ by $24$ from $n$, as all squares are within $6$ 'steps' from the center square. And since there are $49$ squares on the board, that means that the numbers on the board have to be $n-24$ through $n+24$.
But while the $4$ squares at the corners of the sub-board can differ by up to $24$ from $n$, the other squares can differ by at most $20$, and hence there is no way to place the $8$ numbers $n-24$, $n-23$, $n-22$, $n-21$, $n+21$, $n+22$, $n+23$, and $n+24$ on the sub-board. Contradiction!
Obviously, if we can't place different whole numbers on a $7 \times 7$ board with adjacent squares having a difference of at most $4$, then we can;t do it for a $8 \times 8$ board either, since if we could do it for an $8 \times 8$ board, then it would be done for any of the $7 \times 7$ 'sub-boards' as well, which we proved is impossible.