If $u$ subharmonic, why $U_s=\{x\mid u(x)=s\}$ is open?

Question

Let $U$ a domain and $u\in \mathcal C^2(U)\cap \mathcal C^1(\bar U)$ and let $s=\sup_{U}u$. Let $U_s=\{x\mid u(x)=s\}$. I want to prove that $U$ is clopen. I proved that it's closed (because $U=u^{-1}\{s\}$ and $U$ continuous), but I don't understand why for all $x\in U$, there is $r>0$ s.t. $\mathcal B_r(x)\subset U_s$. Any idea ?

Answer 1

Hint

Apply mean value property to $x\mapsto u(x)-s$ that is subharmonic on $U$.

Answer 2

By the maximum principle if $u(x)=s$ then $u(y)=s$ for all $y$ in some disk around $x$.