It does matter if $\Omega$ is finite. I am not sure that you can talk about the support of a measure if $\Omega$ is not metric. So let us assume that $\Omega$ is finite and there is a discrete metric on it, then the support of $\mu$ is simply $\{\omega\in \Omega:\mu(\omega)>0\}$. Moreover, $$\mu(\operatorname{supp}\mu) = \nu(\operatorname{supp}\nu)=1.$$ Again, you should assume that $\mu,\nu$ are probability measure, otherwise your statement is incorrect.
So, $\mu(\Omega) = \nu(\Omega) = 1$ and $\|\mu-\nu\| = \sup\limits_{A\in \mathcal F}|\mu(A)-\nu(A)| =1$ where $\mathcal F = 2^\Omega$. Suppose that there is a point $\omega'$ such that $\mu(\omega')>0$ and $\nu(\omega')>0$, then for any set $A$ $$ |\mu(A)-\nu(A)|<1-\min(\mu(\omega'),\nu(\omega)') $$ which contradicts with the assumption $\|\mu-\nu\|=1$.
Be aware of the following example $\mu$ is measure on $\mathbb R$ with Gaussian density and $\nu = \delta_0$ is concentrated at $\{0\}$. Then $\operatorname{supp}\mu = \mathbb R$ and $\operatorname{supp}\nu = \{0\}$ so $\operatorname{supp}\mu\cap\operatorname{supp}\nu = \{0\}$ and even $\nu(\{0\})=1$. However, still $\|\mu-\nu\|=1$.