Smooth compactly supported functions are dense in C0 on an open Euclidean set

If $\mathrm{\Omega }=\mathrm{\varnothing }$, the zero function suffices. If $n=0$ and $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$, then its singleton supports every function and every function is smooth, so take the approximant $f$ itself. Assume $n\ge 1$. Let $f\in C_0(\mathrm{\Omega })$ and $\epsilon >0$. The set $K:=\{x\in \mathrm{\Omega }:\mathrm{\mid }f(x)\mathrm{\mid }\ge \epsilon \mathrm{/}3\}$ is compact by the definition of vanishing at infinity. If $K=\mathrm{\varnothing }$, the zero function approximates $f$ within $\epsilon \mathrm{/}3$. Otherwise cover $K$ by finitely many inner balls whose doubled closed balls lie in $\mathrm{\Omega }$, using compactness [F1]. Choose the corresponding bumps from [F2] and combine them as $\chi :=1-{\prod }_j(1-{\chi }_j)$. Then $\chi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ and $\chi =1$ on $K$, so $\mathrm{\parallel }f-\chi f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \epsilon \mathrm{/}3$.

Put $g=\chi f$ and $L=\mathrm{supp}g⋐\mathrm{\Omega }$. If $L=\mathrm{\varnothing }$, take the zero approximant. Otherwise, for each $x\in L$, continuity gives $r_x>0$ such that $\bar{B}(x,2r_x)\subset \mathrm{\Omega }$ and $\mathrm{\mid }g(y)-g(x)\mathrm{\mid }<\epsilon \mathrm{/}3$ on $B(x,2r_x)$. Finitely many inner balls $B(x_j,r_j)$ cover $L$. By [F2], let ${\psi }_j$ be nonnegative smooth bumps equal to one on those inner balls and supported in their doubled balls. The sum $s={\sum }_j{\psi }_j$ is positive on a neighborhood of $L$. Repeating the finite cutoff construction of step 1.1 inside the open set $\{s>0\}$ gives $\zeta \in C_c^{\mathrm{\infty }}(\{s>0\})$ with $0\le \zeta \le 1$ and $\zeta =1$ on $L$. Define $a=\zeta s^{-1}{\sum }_{j}g(x_j){\psi }_j$ on $\{s>0\}$ and extend it by zero. Its compact support lies in $\mathrm{\Omega }$, and the extension is smooth because $\mathrm{supp}\zeta ⋐\{s>0\}$. Where $s>0$, the weighted average differs from $g(y)$ by at most $\epsilon \mathrm{/}3$, since each nonzero weight has $\mathrm{\mid }g(x_j)-g(y)\mathrm{\mid }<\epsilon \mathrm{/}3$. Also $\zeta g=g$ everywhere, because $\zeta =1$ on $\mathrm{supp}g$. Therefore $\mathrm{\parallel }a-g{\mathrm{\parallel }}_{\mathrm{\infty }}\le \epsilon \mathrm{/}3$, and step 1.1 gives $\mathrm{\parallel }a-f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2\epsilon \mathrm{/}3<\epsilon$. This uses only finitely many balls, bumps and sampled values, with no integration or infinite selection.