Proof that Poisson formula solves the Neumann Problem for Laplace Equation in Unit Disk

Let $$u: U \mapsto \mathbb{R}$$ be given by $$\begin{gather} u(y) = -\frac{1}{\pi} \int_{\partial U} g(x) \log |x-y|\ \mathrm{d}x \quad \text{with} \quad U := \{ y \in \mathbb{R}^2 : |y| < 1\} \end{gather}$$ which is the Poisson formula for the Neumann problem on the unit disk solving \begin{align} \Delta u &= 0 \quad \text{in} \ \ U\\ \partial_{\nu}u &= g \quad \text{on} \ \ \partial U \end{align} and $$g$$ is continuous on the boundary $$\partial U$$.

Now I want to prove that $$u$$ is harmonic in $$U$$, continuous in $$\overline{U}$$ and that $$\int_{\partial U} u = 0$$ must hold.

I have a hard time understanding how to approach the proof for continuity, because I don't know how to take the value of $$u$$ at the boundary (for a Dirichlet problem $$u=g$$ on $$\partial U$$ and we can take advantage of the continuity of $$g$$). Also, I understand that according to Green's identity there is the consistency condition $$\begin{gather} \int_U \Delta u = \int_{\partial U} \partial_{\nu}u = \int_{\partial U} g = 0 \end{gather}$$ but how can I deduce from this that $$\int_{\partial U} u = 0$$ ? Is it that $$\int_{\partial U} \log |x-y|\ \mathrm{d} x= 1$$? If yes, why? Finally, I want to show that $$\partial_{\nu} u(y) \to g(x)$$ as $$x\to y$$, for $$x \in \partial U$$ and $$y \in U$$, but if I understand all of the above I believe I can proceed with it.

Any hint?