Yes, your bound is optimal and here's how to attain it.
Fix a linear ordering $L$ of $\Omega$. Pick a number $x \in [0,1]$ under the uniform measure and the value of our coupling shall be $(M(x),N(x))$.
Let $M(x)\in \Omega$ be the $L$-greatest $a$ such that $$\mu(\{c\in \Omega: c\le_L a\}\le x.$$
Let $N(x)\in \Omega$ be the $L$-greatest $a$ such that $$\mu(\{c\in \Omega: c\le_L a\}\le x+\frac12\mod 1.$$
This makes $\pi(\Delta)=0$ unless there is a single outcome whose $\mu+\nu>1$, by the way.
I suppose there could be some measurability concerns about $L$ but at least it works in the cases of discrete, or continuous real-valued, random variables.