= Solution
The continuous coefficients are <uniformly continuous> on a compact neighborhood of $\overline W$. For every $x_0\in\overline W$, choose a ball $B_r(x_0)\Subset U$ small enough that
$$
\|a^{ij}-a^{ij}(x_0)\|_{L^\infty(B_r(x_0))}<\frac\theta{2n}
$$
for all $i,j$. Part 4(b), with the frozen <symmetric matrix> $A=(a^{ij}(x_0))$, then applies on this ball.
Choose a finite collection of these balls and a smooth <partition of unity> $(\eta_k)$ that sums to one near $\overline W$, with each $\eta_k$ supported in its corresponding ball. Applying part 4(b) to $\eta_k u$ gives
$$
\|D^2(\eta_k u)\|_2\leq C\|L(\eta_k u)\|_2.
$$
Since $(a^{ij})$ is symmetric, the <Leibniz rule> gives the commutator formula
$$
L(\eta_k u)
=\eta_kLu+2a^{ij}(D_i\eta_k)(D_ju)+a^{ij}(D_{ij}\eta_k)u.
$$
The coefficients and the finitely many derivatives of the <cutoff functions> are bounded, so
$$
\|L(\eta_k u)\|_2
\leq C\left(\|Lu\|_{L^2(U)}+\|u\|_{H^1(U)}\right).
$$
Finally $u=\sum_k\eta_k u$ near $\overline W$. Summing the finite set of local estimates proves
$$
\boxed{\|D^2u\|_{L^2(W)}
\leq C\left(\|Lu\|_{L^2(U)}+\|u\|_{H^1(U)}\right).}
$$
This is the <coefficient-freezing interior second-derivative estimate>.
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