= Solution
There is a small incompatibility in the printed <cutoff function> assumptions. No $\varphi\in C_c^\infty(B_1)$ can equal one on $B_{1/2}$ and satisfy $|\nabla\varphi|\leq2$: for a unit vector $e$, choose $t<1$ beyond its <support>. Continuity gives $\varphi(e/2)=1$, while the <fundamental theorem of calculus> would give $1\leq2(t-1/2)<1$. Replace the number $2$ by a fixed sufficiently large absolute constant $C_0$; such a <cutoff function> exists, and the requested <Caccioppoli inequality> is unchanged apart from its constant. We keep $M=\|\nabla\varphi\|_\infty$ in the calculation, which also yields the formal bound under the printed assumptions.
The <weak formulation> is $\int_{B_1}A\nabla u\cdot\nabla\zeta=0$ for <test functions> $\zeta$. Since $A$ is bounded, this extends by <density of smooth functions in a Sobolev space> to every $\zeta\in H_0^1(B_1)$. Part (a)(ii), followed by <mollification> inside the <compact support>, makes $\varphi^2(u-k)$ an admissible choice. The <product rule> gives
$$
0=\int\varphi^2 A\nabla u\cdot\nabla u
+2\int\varphi(u-k)A\nabla u\cdot\nabla\varphi.
$$
Put $X=\int\varphi^2|\nabla u|^2$ and $Y=\int(u-k)^2|\nabla\varphi|^2$. The lower and upper <uniform ellipticity> bounds, together with <Cauchy-Schwarz inequality>, imply
$$
\lambda X\leq2\Lambda X^{1/2}Y^{1/2}.
$$
If $X=0$ the conclusion is immediate. Otherwise divide by $X^{1/2}$ and square to obtain $X\leq4\Lambda^2\lambda^{-2}Y$. Since $\varphi=1$ on $B_{1/2}$,
$$
\boxed{\int_{B_{1/2}}|\nabla u|^2\,dx
\leq\frac{4\Lambda^2 M^2}{\lambda^2}\int_{B_1}(u-k)^2\,dx.}
$$
For a fixed valid <cutoff function>, $M\leq C_0$, this has the required constant independent of $k$. Formally inserting the printed value $M=2$ gives $16\Lambda^2/\lambda^2$. No derivative of $A$ was taken, so differentiability of the coefficients is unnecessary.
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