Extension by zero for distributions with ambient closed support

The sets $\mathrm{\Omega }$ and ${\mathbb{R}}^n\setminus F$ are open and cover ${\mathbb{R}}^n$ because $F\subseteq \mathrm{\Omega }$. On their overlap $\mathrm{\Omega }\setminus F$, the distribution $u$ vanishes by F2 since its support is contained in $F$. Hence $u$ and the zero distribution on ${\mathbb{R}}^n\setminus F$ are compatible.

F1 glues them to $\tilde{u}$. It restricts to $u$ and is zero on ${\mathbb{R}}^n\setminus F$, so F2 gives support contained in $F$. Any other extension with this support must have the same two restrictions, and uniqueness in F1 makes it equal to $\tilde{u}$.

If $F$ is empty, F2 makes $u=0$ and the extension is zero; if $\mathrm{\Omega }={\mathbb{R}}^n$, the extension is $u$. Ambient closedness ensures that the second member of the cover is open. The proof therefore supplies no extension claim across a boundary when only relative closedness is known. There is no choice use beyond the choice-free sheaf theorem. $\mathrm{\square }$