$C^\infty$ version of Urysohn Lemma in $\Bbb R^n$

Since $K$ is compact and $U^C$ is closed, there is a positive distance, $\Delta$, from $K$ to $U^C$.

We can cover each point $k\in K$ with an open cube $Q_k(\Delta/\sqrt{n})$ centered at $k$ and side $\Delta/\sqrt{n}$. Note that the entire cube is within $\frac\Delta2$ of $k$. Since $K$ is compact, choose a finite subcover of these cubes $\{Q_{k_j}(\Delta/\sqrt{n}):1\le j\le N\}$. For each cube in this subcover, define the function $f_j$ mentioned in step 4 above on $Q_{k_j}(2\Delta/\sqrt{n})$ but divide it by its minimum on $\overline{Q}_{k_j}(\Delta/\sqrt{n})$. Thus, $f_j$ is supported in $Q_{k_j}(2\Delta/\sqrt{n})$ and is $\ge1$ on $Q_{k_j}(\Delta/\sqrt{n})$.

$\sum\limits_{j=1}^Nf_j$ is supported in $U$ and is $\ge1$ on $K$. Then $\phi\circ\sum\limits_{j=1}^Nf_j$ is supported in $U$ and is $1$ on $K$, where

$$\phi(x)=\frac{h_+(x)}{h_+(x)+h_+(1-x)}$$

Note that $\phi\in C^\infty$, and $\phi(x)=0$ for $x\le0$ and $\phi(x)=1$ for $x\ge1$.