[proofplan] Fix the pole $y \in \Omega$ and prove positivity of the function $x \mapsto G(x,y)$ on $\Omega \setminus \{y\}$. Around the pole, remove a small closed ball so that the Green function is harmonic on the remaining punctured domain, vanishes on the outer boundary, and is positive on the inner boundary. The weak maximum principle gives nonnegativity on the punctured domain, and the strong maximum principle upgrades this to strict positivity because the function cannot vanish identically while it is positive on the inner boundary. [/proofplan]

[step:Fix the pole and choose a punctured domain containing the evaluation point] Fix $y \in \Omega$. Define the map $u_y : \Omega \setminus \{y\} \to \mathbb{R}$ by $u_y(x) = G(x,y)$. Let $x_0 \in \Omega \setminus \{y\}$ be arbitrary. Since $\Omega$ is open and $x_0 \neq y$, choose $r > 0$ such that \begin{align*} 0 < r < |x_0 - y| \quad \text{and} \quad \overline{B}(y,r) \subset \Omega. \end{align*} Set \begin{align*} D_r := \Omega \setminus \overline{B}(y,r). \end{align*} Then $x_0 \in D_r$, and by the defining properties of the Dirichlet Green function, $u_y$ is harmonic on $D_r$. [/step]

[step:Use the boundary signs to obtain nonnegativity on the punctured domain]By the Dirichlet boundary condition, $u_y = 0$ continuously on $\partial \Omega$. By the assumed positive fundamental-solution singularity, after decreasing $r > 0$ if necessary while keeping $r < |x_0-y|$ and $\overline{B}(y,r) \subset \Omega$, the function $u_y$ extends continuously to $\partial B(y,r)$ and satisfies $u_y(z) > 0$ for every $z \in \partial B(y,r)$. Therefore $u_y \geq 0$ on the boundary pieces used by the maximum principle for the punctured domain $D_r$, namely on $\partial \Omega \cup \partial B(y,r)$. Since $u_y$ is harmonic on $D_r$, extends continuously to those boundary pieces, and the [maximum principle](/theorems/32) is assumed to apply on $D_r$, it follows that $u_y(x) \geq 0$ for every $x \in D_r$.[/step]

[guided]The purpose of removing $\overline{B}(y,r)$ is to eliminate the singularity of the Green function at the pole. Once the pole is removed, the map $u_y : D_r \to \mathbb{R}$, defined by $u_y(x) = G(x,y)$, is harmonic on the whole domain $D_r = \Omega \setminus \overline{B}(y,r)$. We now check the boundary information needed for the weak maximum principle. On the outer boundary $\partial \Omega$, the Dirichlet condition for the Green function gives $u_y = 0$, with continuous boundary convergence. On the inner boundary $\partial B(y,r)$, the standard singular expansion of the Dirichlet Green function says that $G(\cdot,y)$ has the same leading behaviour as the positive fundamental solution for $-\Delta$ near $y$. Hence, for all sufficiently small $r > 0$ with $\overline{B}(y,r) \subset \Omega$, the function $u_y$ extends continuously to $\partial B(y,r)$ and satisfies $u_y(z) > 0$ for every $z \in \partial B(y,r)$. We chose $r < |x_0-y|$, and decreasing $r$ preserves this inequality, so $x_0$ remains in $D_r$. Thus $u_y \geq 0$ on every boundary component of $D_r$ used by the maximum principle. Since $u_y$ is harmonic on $D_r$, has the required continuous boundary behaviour, and the [maximum principle](/theorems/32) is assumed to apply on this punctured subdomain, the minimum of $u_y$ cannot be negative in the interior. Therefore $u_y(x) \geq 0$ for every $x \in D_r$.[/guided]

[step:Apply the strong maximum principle on the component containing $x_0$]Let $C$ denote the connected component of $D_r$ containing $x_0$. We claim first that $\overline{C} \cap \partial B(y,r) \neq \varnothing$. Since $\Omega$ is a connected open subset of $\mathbb{R}^n$, it is path connected. Choose a continuous path $\gamma : [0,1] \to \Omega$ with $\gamma(0)=x_0$ and $\gamma(1)=y$. Define $t_* := \inf\{t \in [0,1] : \gamma(t) \in \overline{B}(y,r)\}$. Then $0<t_*<1$, $\gamma(t_*) \in \partial B(y,r)$, and $\gamma([0,t_*)) \subset D_r$. Because $\gamma([0,t_*))$ is connected and contains $x_0$, it is contained in $C$, so $\gamma(t_*) \in \overline{C} \cap \partial B(y,r)$. Now suppose, for contradiction, that $u_y(x_0)=0$. Since $u_y \geq 0$ on $D_r$, the point $x_0$ is an interior minimum of the harmonic function $u_y|_C$ on the connected domain $C$. The set $C$ is an open connected subdomain of the punctured domain $D_r$, and the restriction $u_y|_C : C \to \mathbb{R}$ is harmonic. Therefore the assumed punctured-domain [strong maximum principle](/theorems/32), applied to this connected component, implies that $u_y|_C$ is constant on $C$. This constant is $0$, because $u_y(x_0)=0$. Choose $z_* \in \overline{C} \cap \partial B(y,r)$. Since $u_y$ extends continuously to $\partial B(y,r)$ and $u_y=0$ on $C$, continuity along points of $C$ approaching $z_*$ gives $u_y(z_*)=0$. This contradicts the already established boundary sign $u_y(z_*)>0$ on $\partial B(y,r)$. Therefore $u_y(x_0)>0$.[/step]

[guided]The strong maximum principle must be used on a connected domain, so we first isolate the correct component. Let $C$ be the connected component of $D_r = \Omega \setminus \overline{B}(y,r)$ containing $x_0$. The key topological point is that $C$ really reaches the inner boundary sphere. Since $\Omega$ is open and connected in $\mathbb{R}^n$, it is path connected. Choose a continuous path $\gamma : [0,1] \to \Omega$ with $\gamma(0)=x_0$ and $\gamma(1)=y$. The path starts outside $\overline{B}(y,r)$ because $r<|x_0-y|$, and it ends inside that closed ball because $y \in B(y,r)$. Define $t_* := \inf\{t \in [0,1] : \gamma(t) \in \overline{B}(y,r)\}$. By continuity of $\gamma$, we have $0<t_*<1$ and $\gamma(t_*) \in \partial B(y,r)$; before time $t_*$, the path lies in $D_r$. The set $\gamma([0,t_*))$ is connected and contains $x_0$, so it lies in the connected component $C$. Hence $\gamma(t_*)$ is a limit point of $C$ lying on the inner sphere, and therefore $\overline{C} \cap \partial B(y,r) \neq \varnothing$. Now assume toward a contradiction that $u_y(x_0)=0$. From the weak maximum principle step we already know $u_y \geq 0$ on $D_r$, so $x_0$ is an interior minimum of the harmonic function $u_y|_C : C \to \mathbb{R}$. The [strong maximum principle](/theorems/32) applies to the connected component $C$, not merely to the possibly disconnected set $D_r$. Indeed, $C$ is an open connected subdomain of $D_r$, the restriction $u_y|_C : C \to \mathbb{R}$ is harmonic because $u_y$ is harmonic on $D_r$, and $x_0 \in C$ is an interior point where the nonnegative harmonic function attains its minimum value $0$. Thus the assumed punctured-domain strong maximum principle may be applied after restricting to this connected component. Therefore $u_y|_C$ is constant on $C$. Since the value at $x_0$ is $0$, the constant is $0$. Choose $z_* \in \overline{C} \cap \partial B(y,r)$. The [boundary regularity](/theorems/99) already used in the weak maximum principle step gives continuous extension of $u_y$ to the inner boundary sphere. Since $u_y$ is identically $0$ on $C$, taking any sequence of points of $C$ converging to $z_*$ gives $u_y(z_*)=0$. But the inner boundary was chosen so that $u_y(z)>0$ for every $z \in \partial B(y,r)$, in particular $u_y(z_*)>0$. This contradiction shows that the assumption $u_y(x_0)=0$ was impossible, so $u_y(x_0)>0$.[/guided]

[step:Conclude positivity for all distinct pairs] The point $x_0 \in \Omega \setminus \{y\}$ was arbitrary. Therefore $G(x_0,y) = u_y(x_0) > 0$ for every $x_0 \in \Omega \setminus \{y\}$. Since $y \in \Omega$ was arbitrary, it follows that $G(x,y) > 0$ for all $x,y \in \Omega$ with $x \neq y$. [/step]