**Step 1: The Inequality for Smooth Functions** We first establish the inequality for functions with compact support. Let $\phi \in C_c^\infty(U)$. We extend $\phi$ to be zero on $\mathbb{R}^n \setminus U$. Since $U$ is contained in the slab $a < x_n < a+d$, for any point $x = (x', x_n) \in U$, we can express $\phi(x)$ using the Fundamental Theorem of Calculus along the $x_n$ direction, starting from the boundary $x_n = a$ (where $\phi = 0$): \begin{align*} \phi(x', x_n) = \int_a^{x_n} \partial_{x_n} \phi(x', t) \, dt. \end{align*} We estimate the absolute value: \begin{align*} |\phi(x', x_n)| \le \int_a^{x_n} |\partial_{x_n} \phi(x', t)| \, dt \le \int_a^{a+d} |\partial_{x_n} \phi(x', t)| \, dt. \end{align*} If $p=1$, the inequality follows immediately. Assume $p > 1$ and let $q$ be the conjugate exponent ($1/p + 1/q = 1$). We apply Hölder's inequality to the integral on the right: \begin{align*} \int_a^{a+d} 1 \cdot |\partial_{x_n} \phi(x', t)| \, dt &\le \left( \int_a^{a+d} 1^q \, dt \right)^{1/q} \left( \int_a^{a+d} |\partial_{x_n} \phi(x', t)|^p \, dt \right)^{1/p} \\ &= d^{1/q} \left( \int_a^{a+d} |\partial_{x_n} \phi(x', t)|^p \, dt \right)^{1/p}. \end{align*} Substituting this back into the estimate for $|\phi|$ and raising both sides to the power $p$: \begin{align*} |\phi(x', x_n)|^p \le d^{p/q} \int_a^{a+d} |\partial_{x_n} \phi(x', t)|^p \, dt. \end{align*} We now integrate this inequality over the full domain $U$. We perform the $x_n$ integration first. Notice that the right-hand side does not depend on $x_n$: \begin{align*} \int_a^{a+d} |\phi(x', x_n)|^p \, dx_n &\le \int_a^{a+d} \left( d^{p/q} \int_a^{a+d} |\partial_{x_n} \phi(x', t)|^p \, dt \right) \, dx_n \\ &= d \cdot d^{p/q} \int_a^{a+d} |\partial_{x_n} \phi(x', t)|^p \, dt \\ &= d^p \int_a^{a+d} |\partial_{x_n} \phi(x', t)|^p \, dt. \end{align*} (Note: We used $1 + p/q = p$). Finally, integrating over the remaining variables $x'$: \begin{align*} \int_{\mathbb{R}^{n-1}} \int_a^{a+d} |\phi|^p \, dx_n \, dx' &\le d^p \int_{\mathbb{R}^{n-1}} \int_a^{a+d} |\partial_{x_n} \phi|^p \, dt \, dx'. \end{align*} This simplifies to $\|\phi\|_{L^p(U)}^p \le d^p \|\partial_{x_n} \phi\|_{L^p(U)}^p$. Taking the $p$-th root and noting $|\partial_{x_n} \phi| \le |\nabla \phi|$: \begin{align*} \|\phi\|_{L^p(U)} \le d \|\nabla \phi\|_{L^p(U)}. \end{align*}

**Step 2: Extension to $W_0^{1,p}(U)$ via Density** Let $u$ be an arbitrary element of $W_0^{1,p}(U)$. By the definition of $W_0^{1,p}(U)$ as the closure of $C_c^\infty(U)$ under the $W^{1,p}$ norm, there exists a sequence of smooth functions $\{\phi_k\}_{k=1}^\infty \subset C_c^\infty(U)$ such that $\phi_k \to u$ strongly in $W^{1,p}(U)$. This convergence implies convergence of the norms: 1. $\|\phi_k\|_{L^p(U)} \to \|u\|_{L^p(U)}$. 2. $\|\nabla \phi_k\|_{L^p(U)} \to \|\nabla u\|_{L^p(U)}$.

From Step 1, the inequality holds for every element of the sequence: \begin{align*} \|\phi_k\|_{L^p(U)} \le d \|\nabla \phi_k\|_{L^p(U)}. \end{align*} Taking the limit as $k \to \infty$ on both sides: \begin{align*} \lim_{k \to \infty} \|\phi_k\|_{L^p(U)} &\le d \lim_{k \to \infty} \|\nabla \phi_k\|_{L^p(U)} \\ \|u\|_{L^p(U)} &\le d \|\nabla u\|_{L^p(U)}. \end{align*} Thus, the inequality holds for all $u \in W_0^{1,p}(U)$.