If $u \in H^1(U)$, then $Du=0$ a.e. on the set $\{u=0\}$ 
Provide details for the following alternative proof that if $u \in H^1(U)$, then $$Du = 0 \text{ a.e. } \, \text{ on the set } \{u=0\}. $$
  Let $\phi$ be a smooth, bounded and nondecreasing function, such that $\phi'$ is bounded and $\phi(z)=z$ if $|z| \le 1$. Set $$u^\epsilon(x) := \epsilon \phi(u/\epsilon).$$ Show that $u^\epsilon \rightharpoonup 0$ weakly in $H^1(U)$ and therefore $$\int_U Du^\epsilon \cdot Du \, dx = \int_U \phi'(u/\epsilon) |Du|^2 \, dx \to 0.$$ Employ this observation to finish the proof.

PDE by Evans, 2nd edition: Chapter 5, Exercise 19
Here is my work so far. I tried what I can; are there any errors in this?
As $\epsilon \to 0$, 
\begin{align}
\|u^\epsilon\|_{H^1(U)} &:= \|u^\epsilon \|_{L^2(U)} + \|D u^\epsilon \|_{L^2(U)} \\
&= \left\|\epsilon \phi\left(\frac u{\epsilon} \right) \right\|_{L^2(U)}+\|\phi'\left(\frac u{\epsilon} \right) Du \|_{L^2(U)} \\
&\le \left\|C \epsilon \frac u{\epsilon} \right\|_{L^2(U)} + \left\|C \cdot Du \right\|_{L^2(U)} \\
&\le C(\left\| u \right\|_{L^2(U)} + \left\| Du \right\|_{L^2(U)}). 
\end{align}
Hence, $u^\epsilon \rightharpoonup 0$ weakly in $H^1(U)$. Therefore, at $u=0$,
$$\require{cancel} \int_U \left. \cancelto{1}{\phi'(0/\epsilon)} |Du|^2  \right\vert_{u=0}\, dx \to 0$$
 gives $\int_U \left. |Du|^2 \right\vert_{u=0} \, dx \to 0$. Hence, $\left. Du \right\vert_{u=0} =0$ a.e., or $Du=0$ a.e. on the set $\{u=0\}$.
 A: Shown $u^\epsilon\to 0$ weakly in $L^2$ is the easy part, I think you already proved it. (I looked at comments, it has a good hint)
The trick part is to show $\nabla u^\epsilon\to 0$ weakly in $L^2$ as well. Applying chain rule will lead you nowhere. You need following result in Functional analysis

Let $X$ be a Banach space, $S$ be a total subset of $X^*$, i.e., the span of $S$ is dense in $X^*$, $(x_n)$ be a sequence in $X$ and $x\in X$ is given. Then $x_n\to x$ weakly as $n\to\infty$ if and only if $({x_n})$ is bounded and 
$$
\left<{y^*,x_n}\right>\to\left<{y^*,x}\right>\text{ for all }y^*\in S.
$$

Hence, by definition we need to prove that 
$$ \int_U \partial_i u^\epsilon\,v\,dx\to  0\tag w$$
for all $v\in L^2(U)$, but now we only need to prove that 
$$ \int_U \partial_i u^\epsilon\,v\,dx\to  0\tag 1$$
for all $v\in C_c^\infty(U)$
and hence you can use IBP and have $(1)$ is equivalent to
$$ \int_U  u^\epsilon\,\partial_i v\,dx\to  0$$
which is true since $u^\epsilon\to 0$ weakly in $L^2$.
