Let $u\in\mathscr C^2(U)\cap\mathscr C(\bar U)$ be harmonic in the non-empty open and connected set $U\subset\mathbb R^n$. If there exists a Point $x_0\in U$, so that $u$ has a local Maximum at $x_0$, then $u$ is constant in $U$.
Proof:
Let $x_0$ be an point in $U$ with $u(x_0)\geq u(x)$ for all $x$ in some neighbourhood of $x_0$. Without loss of generality we assume the neighbourhood to be $\bar{B}_R(x_0)$ for some $R>0$. Due to the mean value property, we have: $$\frac1{\omega_nR^n}\int_{B_R(x_0)}u(x_0)dx=u(x_0)=\frac1{\omega_nR^n}\int_{B_R(x_0)}u(x)dx \Rightarrow \frac1{\omega_nR^n}\int_{B_R(x_0)}u(x_0)-u(x)dx=0$$. $u(x_0)-u(x)\geq0$ in $B_R(x_0)$ implies $u(x_0)-u(x)=0$ in $\bar B_R(x_0)$. We conclude that the set $V=\{x\in U|u(x)=u(x_0)\}$ is open in $U$. Since $V$ is clearly also closed, we have $U=V$ due to the connectedness.
I don't understand why $V$ is open in $U$. Help would be greatly appreciated as my colleage and I are on the verge of a mental breakdown.