Ideed, because $$ v_k\to u \;\; \mbox{in}\;\;W^{1,p}(U),$$ $$ w_k \to u \;\; \mbox{in} \;\; W^{2,p}(U) $$ we can suppose that $$D v_k\to D u, $$ $$ D w_k \to D u$$ pointwise. Then, pointwise we have $$D v_k \cdot D w_k \vert D w_k \vert ^ {p-2} \to \vert Du \vert ^p\in L^1. $$ By Young’s inequality associated to Swhartz’s Inequality $$\vert D v_k \cdot D w_k \vert D w_k \vert ^ {p-2}\vert \le C (\vert Dw_k \vert^p+\vert Dv_k \vert^p)$$ and as $$\int_U (\vert Dw_k \vert^p+\vert Dv_k \vert^p) \to 2\int_U \vert Du\vert^p,$$ the General Lebesgue Dominated Theorem implies that
$$\displaystyle \int_{U} D v_k \cdot D w_k \vert D w_k \vert ^ {p-2} \to \int_{U} \vert D u \vert ^p.$$
$A\subset\subset B$ is read "$A$ is compactly contained in $B$", sometimes also written with \Subset$\Subset$. According to Evans (2nd ed., pg 698), it means that, for open sets $A,B$, we have $A\subset \overline A\subset B$ and that $\overline A$ is compact:
(vi) $U, V,$ and $W$ usually denote open subsets of $\mathbb{R}^{n} .$ We write
$$
V \subset \subset U
$$
if $V \subset \bar{V} \subset U$ and $\bar{V}$ is compact, and say $V$ is compactly contained in $U$.
As $V\Subset U \subset \mathbb R^n$, in particular $\overline V\subset U$. I believe $U$ is (as in the above) implicitly an open set; thus $d(\overline V,U^c) \in (0,\infty]$. Set $l = \min(1,d(\overline V,U^c))\in(0,1]$.
For each $x\in \overline V$, let $B_x := \mathbb B(x,l/2) $ be the open ball of radius $l/2$ around $x$. Clearly $\{B_x\}_{x\in \overline V}$ is a cover of $\overline V$, so by compactness there is a finite number of them, $B_1,\dots,B_N$ that cover $\overline V$. Now we can take
$W:= B_1\cup \dots\cup B_N$. Since each $\overline{B_i}$ is a subset of $U$ (there is no intersection with $U^c$), $\overline W\subset U$. $\overline W$ is a finite union of closed and bounded balls; hence it is closed and bounded in $\mathbb R^n$, and hence compact. Summary: we created a second open subset $W$ of $U$ such that $V\Subset W \Subset U$. Written using normal subset notation, we have
$$ V\subset \overline V \subset W \subset \overline W \subset U.$$
The fact that $\overline W$ is compact immediately implies the uniform continuity of $f$ on $\overline W$, and hence on $W$.
You might ask why we needed $W$ in the first place since $V$ is already precompact; the answer is that we need some space in order to mollify and get a function defined on $V$, for all $\epsilon$ sufficiently small.
Best Answer
Use that $|B(x,\epsilon)| = |B(0,1)|\epsilon^n$ and that $\eta(\frac{x-y}{\epsilon}) \leq C$.