= Solution
A local form of the <Harnack inequality for harmonic functions> is: if $u\geq0$ is harmonic on a neighbourhood of $\overline{B_{4r}(y)}$, then
$$
\boxed{\sup_{B_r(y)}u\leq3^n\inf_{B_r(y)}u.}
$$
For $x,z\in B_r(y)$, the triangle inequality gives $B_r(x)\subset B_{3r}(z)\subset B_{4r}(y)$. The ball <mean value property for harmonic functions> and nonnegativity therefore imply
$$
u(x)=\frac1{|B_r|}\int_{B_r(x)}u\leq\frac1{|B_r|}\int_{B_{3r}(z)}u=\frac{|B_{3r}|}{|B_r|}u(z)=3^nu(z).
$$
Taking the supremum over $x$ and infimum over $z$ proves the inequality, including the case where the infimum is zero. No strict positivity assumption was used.
On a connected domain $\Omega$, the equivalent compact-set statement is that for each nonempty compact $K\Subset\Omega$ there exists $C(K,\Omega,n)$ with
$$
\boxed{\sup_Ku\leq C(K,\Omega,n)\inf_Ku.}
$$
Here is the passage from balls to this statement. Cover $K$ by finitely many small balls $B_{r_j}(y_j)$ whose quadrupled closed balls lie in $\Omega$. Join their centres to a fixed centre by paths in $\Omega$. Each of the finitely many paths has compact image and positive distance from the boundary, so subdividing it gives a finite chain of overlapping small balls whose quadrupled closures also lie in $\Omega$. The local inequality compares any two points in each ball; using a point in each successive overlap propagates comparisons along the chain. The number of comparisons is bounded uniformly over this finite collection of chains and covering balls. Multiplying the factors $3^n$ gives one finite constant for all pairs $x,z\in K$, proving the compact-set inequality. In particular a nonnegative harmonic function vanishing at one interior point vanishes throughout a connected domain.
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