Idea of the proof: Work directly on the whole domain $U$ (no puncturing). Pair the distribution $-\Delta_y G(x,\cdot)=\delta_x$ with $u$, integrate by parts once (Green’s identity on $U$), and use that $-\Delta u = f$ in $U$ and $G(x,\cdot)=0$ on $\partial U$. The inner product with $\delta_x$ contributes $u(x)$; all other pieces produce exactly the two integrals in the formula.

**Step 1 (start from the distributional identity).** For fixed $x \in U$, the assumption $-\Delta_y G(x,\cdot)=\delta_x$ in $\mathcal{D}'(U)$ means that for every $\varphi \in C_c^\infty(U)$, \begin{align*} \varphi(x) \;=\; \langle \delta_x,\varphi\rangle \;=\; \langle -\Delta_y G(x,\cdot), \varphi \rangle \;=\; \int_{U} G(x,y)\, \Delta \varphi(y)\, d\mathcal{L}^n(y). \tag{1} \end{align*}

**Step 2 (replace the test function by $u$ via Green’s identity on $U$).** Because $u \in C^2(U)\cap C(\overline{U})$ and $G(x,\cdot)\in L^1_{\mathrm{loc}}(U)$ with the stated traces, Green’s identity on $U$ holds (in the classical sense for $u$ and in the distributional sense for $G$): \begin{align*} \int_{U} \big( u(y)\,\Delta_y G(x,y) - G(x,y)\,\Delta u(y) \big) \, d\mathcal{L}^n(y) = \int_{\partial U} \big( u(y)\,\partial_\nu G(x,y) - G(x,y)\,\partial_\nu u(y) \big) \, d\mathcal{H}^{n-1}(y). \tag{2} \end{align*} Each term is meaningful by the assumptions (trace of $G$, integrability of $\partial_\nu G$, and continuity of $u,\partial_\nu u$).

**Step 3 (insert the equations for $G$ and $u$ and identify the terms).** Since $-\Delta_y G(x,\cdot)=\delta_x$ and $-\Delta u = f$ in $U$, and $G(x,\cdot)=0$ on $\partial U$, we compute: \begin{align*} \text{LHS of (2)} &= \int_U u(y)\, \Delta_y G(x,y) \, d\mathcal{L}^n(y) \;-\; \int_U G(x,y)\, \Delta u(y) \, d\mathcal{L}^n(y) \\ &= - \,\langle \delta_x, u \rangle \;+\; \int_U f(y)\, G(x,y) \, d\mathcal{L}^n(y) \\ &= -\, u(x) \;+\; \int_U f(y)\, G(x,y) \, d\mathcal{L}^n(y), \tag{3} \end{align*} and \begin{align*} \text{RHS of (2)} &= \int_{\partial U} u(y)\,\partial_\nu G(x,y)\, d\mathcal{H}^{n-1}(y) \;-\; \underbrace{\int_{\partial U} G(x,y)\,\partial_\nu u(y)\, d\mathcal{H}^{n-1}(y)}_{=\,0 \text{ since } G(x,\cdot)=0 \text{ on } \partial U}} \\ &= \int_{\partial U} g(y)\,\partial_\nu G(x,y)\, d\mathcal{H}^{n-1}(y). \tag{4} \end{align*}

**Step 4 (equate the two sides of Green’s identity).** From (2)–(4) we obtain \begin{align*} -\, u(x) \;+\; \int_U f(y)\, G(x,y)\, d\mathcal{L}^n(y) = \int_{\partial U} g(y)\,\partial_\nu G(x,y)\, d\mathcal{H}^{n-1}(y), \end{align*} which is equivalent to \begin{align*} u(x) = - \int_{\partial U} g(y)\,\partial_\nu G(x,y)\, d\mathcal{H}^{n-1}(y) \;+\; \int_{U} f(y)\, G(x,y)\, d\mathcal{L}^{n}(y), \end{align*} as claimed.