Idea of the proof: Choose a fixed interior scale $r > 0$ depending only on $V$ (distance from $V$ to the boundary). First prove a **local comparison**: if $x,y \in V$ with $|x-y| \le r$, then $u(x)$ and $u(y)$ differ by at most the factor $2^{n}$. This is obtained by (i) using the mean-value identity at $x$, (ii) enlarging the averaging ball at $x$ from radius $r$ to $2r$ (only changing the normalizing volume), and (iii) using $u \ge 0$ to replace the integral over $B(x,2r)$ by the integral over the included ball $B(y,r)$. Then connect arbitrary $x,y \in V$ by a finite chain of such short hops; iterating the local comparison gives the desired global constant.

**Step 1 (fix an interior radius; why it is positive).** Let \begin{align*} r := \tfrac{1}{4}\,\mathrm{dist}(V,\partial U). \end{align*} Because $V \Subset U$, the distance $\mathrm{dist}(V,\partial U)$ is strictly positive; hence $r > 0$ and, for every $z \in V$, $B(z,2r) \subset U$.

**Step 2 (local comparison when $|x-y| \le r$, done in two substeps).** Fix $x,y \in V$ with $|x-y| \le r$. Recall that by translation invariance of Lebesgue measure, the volume of a ball depends only on its radius, so $\mathcal{L}^n(B(x,\rho)) = \mathcal{L}^n(B(0,\rho))$ for any center.

Step 2.a (enlarge the averaging ball but only change the normalizing constant): \begin{align*} u(x) &= \frac{1}{\mathcal{L}^n(B(0,r))} \int_{B(x,r)} u(z) \, d\mathcal{L}^n(z) \\ &\ge \frac{1}{\mathcal{L}^n(B(0,2r))} \int_{B(x,r)} u(z) \, d\mathcal{L}^n(z), \end{align*} since $\mathcal{L}^n(B(0,2r)) \ge \mathcal{L}^n(B(0,r))$ and $u \ge 0$.

Step 2.b (replace the domain of integration using inclusion and nonnegativity): because $|x-y| \le r$, we have $B(y,r) \subset B(x,2r)$, hence \begin{align*} \int_{B(x,2r)} u \, d\mathcal{L}^n \;\ge\; \int_{B(y,r)} u \, d\mathcal{L}^n. \end{align*} Combining with Step 2.a and inserting the intermediate integral yields \begin{align*} u(x) &\ge \frac{1}{\mathcal{L}^n(B(0,2r))} \int_{B(x,2r)} u(z) \, d\mathcal{L}^n(z) \ge \frac{1}{\mathcal{L}^n(B(0,2r))} \int_{B(y,r)} u(z) \, d\mathcal{L}^n(z) \\ &= \frac{\mathcal{L}^n(B(0,r))}{\mathcal{L}^n(B(0,2r))} \cdot \frac{1}{\mathcal{L}^n(B(0,r))} \int_{B(y,r)} u(z) \, d\mathcal{L}^n(z) \\ &= 2^{-n} \, u(y), \end{align*} because $\mathcal{L}^n(B(0,2r)) = 2^n \mathcal{L}^n(B(0,r))$ and the mean-value property gives $u(y) = \tfrac{1}{\mathcal{L}^n(B(0,r))} \int_{B(y,r)} u$. Exchanging $x$ and $y$ we also get $u(y) \ge 2^{-n} u(x)$. Therefore, \begin{align*} 2^{-n} u(y) \le u(x) \le 2^{n} u(y) \qquad \text{whenever } |x-y| \le r. \tag{LC} \end{align*}

**Step 3 (finite chain inside $V$).** Since $V$ is connected and $\overline{V}$ is compact, it is path-connected, and there exists a finite $N=N(V)$ and points $x=x_0,x_1,\dots,x_N=y$ in $V$ with $|x_{j+1}-x_j| \le r$ for $j=0,\dots,N-1$. (Construct a finite cover of $\overline{V}$ by balls of radius $r/2$ and follow any path from $x$ to $y$ through this finite cover.)

**Step 4 (iterate the local estimate and conclude).** Applying (LC) along the chain, \begin{align*} u(x_0) \le 2^{n} u(x_1) \le 2^{2n} u(x_2) \le \cdots \le 2^{nN(V)} u(x_N), \end{align*} and reversing the roles gives the two-sided bound with $C := 2^{nN(V)}$. This depends only on $n$ and on $V$ (through $r$ and the covering number), completing the proof.