Suppose this was wrong. Then for some $C$, we have $\vert a_{i+1}-a_i\vert \le 2d$ as well as $\vert a_i-i\vert \ge d$ for all $i \ge C$. So either $a_M \le M-d$ for some $M \ge C$ or $a_n \ge n+d$ for all $n \ge C$. In the first case we have $a_{M+1} \le a_M+2d \le M+d<M+d+1$ and hence $a_{M+1} \le (M+1)-d$. By induction, the same argument shows that $a_i \le i-d$ for all $i \ge M$. But then for large $N$, we have $a_k \le N-d$ for all $k \le N$. Hence these $N$ values of the sequence can only take $N-d$ possible values which contradicts the sequence being a permutation. Similarly, if $a_n \ge n+d$ for all $n \ge C$, then only the first $C$ elements of the sequence can take values at most $C+d$ which again contradicts the sequence being a permutation.