[proofplan] We show that $T$ vanishes on the union $\bigcup_{i \in I} U_i$ by reducing to a finite sub-union via compactness. Given a test function $\varphi$ supported in the union, its compact support is covered by finitely many $U_i$. A partition of unity subordinate to this finite cover decomposes $\varphi$ into pieces, each supported in a single $U_i$ where $T$ vanishes. [/proofplan] [step:Extract a finite subcover and partition of unity] Let $\varphi \in \mathcal{D}(\bigcup_{i \in I} U_i)$. The set $K := \mathrm{supp}(\varphi)$ is compact and contained in $\bigcup_{i \in I} U_i$. By compactness, there exist finitely many indices $i_1, \ldots, i_m \in I$ such that $K \subseteq U_{i_1} \cup \cdots \cup U_{i_m}$. By the [Existence of Partitions of Unity](/theorems/57), there exist $\chi_1, \ldots, \chi_m \in C^\infty(\Omega)$ satisfying: $\mathrm{supp}(\chi_j) \subset U_{i_j}$, $0 \leq \chi_j \leq 1$, and $\sum_{j=1}^m \chi_j = 1$ on a neighbourhood of $K$. [/step] [step:Decompose $\varphi$ and conclude] Since $\sum_{j=1}^m \chi_j = 1$ on a neighbourhood of $\mathrm{supp}(\varphi)$, we have \begin{align*} \varphi &= \sum_{j=1}^m \chi_j \varphi. \end{align*} Each product $\chi_j \varphi$ belongs to $\mathcal{D}(U_{i_j})$: it is smooth, and $\mathrm{supp}(\chi_j \varphi) \subseteq \mathrm{supp}(\chi_j) \cap \mathrm{supp}(\varphi) \subseteq U_{i_j} \cap K$, which is compact. By hypothesis, $T$ vanishes on each $U_{i_j}$, so $T(\chi_j \varphi) = 0$ for each $j$. By linearity: \begin{align*} T(\varphi) &= \sum_{j=1}^m T(\chi_j \varphi) = 0. \end{align*} [/step]