Well, this was just the intuition for the proof. It is not needed. But what I meant is that if $\hat{\pi}$ is any optimal coupling for the monotonic distribution, then we should have $\hat{\pi}_{(i,i),(i',j')} = 0$ if $||(i,i)-(i',j')||_1 > ||(i',i')-(i',j')||_1$. I.e. if we cannot go from (i,i) to (i',j') by always going away from the diagonal, then $\hat{\pi}_{(i,i),(i',j')}$ should be zero.

The coupling $\pi$ is only optimal for the monotonic/comonotonic case. As you mentioned, this only works with the $l_1$ norm. For the $l_1$ norm, we can visualize going from the diagonal to the other lattice points in $\{1,...,N\}^2$ in horizontal/vertical steps, and the $l_1$ norm is just the number of steps. The idea was, if we are going over from the monotonic distribution to the uniform distribution, any coupling that moves mass strictly away from the diagonal is going to be optimal, and the simplest of those is just $\pi$.