A Euclidean bump for a compact set inside an open set

If $K=\mathrm{\varnothing }$, the zero function has empty support and satisfies the claim. Suppose henceforth that $K\mathrm{\ne }\mathrm{\varnothing }$. Consider all triples $(p,r,R)$ with $p\in K$, $0<r<R$, and $B_R(p)\subseteq U$. Their inner balls $B_r(p)$ form an indexed open cover of $K$: openness gives a ball $B_R(p)\subseteq U$ at each $p$, and $r=R\mathrm{/}2$ is admissible. This defines the whole family without making a simultaneous choice at every point.

By [L1], finitely many admissible triples $(p_i,r_i,R_i)$, $1\le i\le m$, have inner balls covering $K$. For each of these finitely many triples use the explicit formula in [L2], with $x$ replaced by $x-p_i$, to define ${\rho }_i$. Translation preserves the coordinate-derivative formulas, so each ${\rho }_i$ is smooth, lies in $[0,1]$, equals $1$ on ${\bar{B}}_{r_i}(p_i)$, and has support contained in $B_{R_i}(p_i)\subseteq U$.

Define $\rho =1-{\prod }_{i=1}^m(1-{\rho }_i)$. Repeated use of [L3] shows that each iterated coordinate derivative is a finite sum of products of continuous derivatives of the ${\rho }_i$, so $\rho$ is smooth. Since every factor lies in $[0,1]$, so does $\rho$. On each inner ball one factor is zero; hence $\rho =1$ on their union, an open neighbourhood of $K$.

Let $F={\bigcup }_{i=1}^m\mathrm{supp}{\rho }_i$. Each support is closed by [F1], so the finite union $F$ is closed, and step 2.1 gives $F\subseteq U$. Off $F$, every ${\rho }_i$ vanishes, so the formula in step 3.1 gives $\rho =0$. Therefore the closure of the nonzero locus of $\rho$ lies in $F$: $\mathrm{supp}\rho \subseteq F\subseteq U$.

Steps 3.1 and 4.1 give all three required properties; step 1.1 handled the empty case.