For each given $u \in {\scr D}'(\Omega)$ and $f \in {\scr D}(\Omega)$:
${\rm supp}\ u \ \cap \ {\rm supp }\ f = \emptyset \implies u(f) =0 $
The proof uses the partition of unity. My issue is... what is there to prove? The reason probably has to deal with the fact that both ${\rm supp}\ u \ $and $ {\rm supp}\ f$ are closed, while working with a partition (made by open sets) solves problems that can occur at the boundary of the intersection. However, since such boundary is void (as a direct consequence of the hypothesis), I don't understand the necessity of all this effort.
You have a class of open sets $U_\alpha$ on which $u$ vanishes, i.e. $\langle u, \varphi \rangle = 0$ forall $\varphi\in\mathcal{D}(U_\alpha)$.
Now consider some $\varphi \in \mathcal{D}(\bigcup_\alpha U_\alpha).$ The support of $\varphi$ might cover several of the $U_\alpha$'s. It is not trivial then that $\langle u, \varphi \rangle = 0.$ Therefore $\varphi$ has to be written as a sum of several (finitely many!) $\varphi_k \in \mathcal{D}(U_{\alpha_k})$ for which $\langle u, \varphi_k \rangle = 0$ by definition: $$ \langle u, \varphi \rangle = \langle u, \sum_k \varphi_k \rangle = \sum_k \langle u, \varphi_k \rangle = \sum_k 0 = 0. $$