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.
Let your random variable $M$ realizing $\mu$ be defined as follows. $M(x)\in \Omega$ is the $L$-greatest $a$ such that $$\mu(\{c\in \Omega: c\le_L a\}\le x.$$
Similarly for $\nu$ with $x+1/2$ mod 1: Let your random variable realizing $\nu$ be as follows. Your outcome in $\Omega$ is the $L$-greatest $b$ such that the $\nu$ probability of an outcome $L$-below $b$ is $\le (x+1/2$ 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.