How to prove the Extension Theorem of $ {{W}^{1,p}}\left( U \right) $ when $ p= \infty $?

Question

I am studying partial differential equations from Mr. Evans' book. In Chapter 5, Section 5.4, I learned the extension theorem:

Theorem Assume $ U $ is bounded and $\partial U$ is $C^1$. Select a bounded open set $V$ such that $U\subset \subset V$(which means $U\subset \overline{U}\subset V$ and $\overline{U}$ is compact). Then there exists a bounded linear operator $$E:{{W}^{1,p}}\left( U \right)\to {{W}^{1,p}}\left( {{\mathbb{R}}^{n}} \right)$$ such that for each $u\in {{W}^{1,p}}\left( U \right)$:

(i)$Eu=u\ \text{a}\text{.e}\text{. in}\ U,$

(ii)$Eu$ has support within $V$,

(iii)${{\left\| Eu \right\|}_{{{W}^{1,p}}\left( {{\mathbb{R}}^{n}} \right)}}\le C{{\left\| u \right\|}_{{{W}^{1,p}}\left( U \right)}}$, the constant $C$ depending only on $p,U,V$.

The theorem's condition allows $ p= \infty $, but the theorem actually only proves the case $ p < \infty$.

At the end of the proof, it says, "The case $ p= \infty $ is left as an exercise", but I don't know how to prove it.

I think the original proof works for $1\leq p \leq \infty$ and does not exclude the case when $p=\infty$. Can you provide a screenshot of your copy? — Tab1e, Sep 16 '21 at 08:10
The proof as given fails at $p=\infty$ because it uses the density of $C^\infty(\overline U)$ in $W^{1,p}(U)$ @Tab1e — Calvin Khor, Sep 16 '21 at 11:33
@Tab1e I checked both editions of Evans, the remark about $p=\infty$ was only added in the 2nd edition. Some links about the closure of smooth functions in Lipschitz/Holder spaces: link1, link2, link3 (since $W^{1,\infty}=\text{Lip}$) — Calvin Khor, Sep 16 '21 at 11:45

How to prove the Extension Theorem of $ {{W}^{1,p}}\left( U \right) $ when $ p= \infty $?

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