Maximum principle for harmonic functions

Suppose $u$ attains its maximum value $M$ over $\overline{\Omega}$ at an interior point $x_0 \in \Omega$. Let $S = \{x \in \Omega : u(x) = M\}$; by continuity $S$ is closed in $\Omega$, and it is nonempty since $x_0 \in S$.

$S$ is also open: for any $x_1 \in S$, choose a small disk $B_r(x_1) \subset \Omega$. The mean value property gives $M = u(x_1) = \frac{1}{2\pi}\int_0^{2\pi} u(x_1 + r\cos\theta, x_1 + r\sin\theta)\, d\theta$, an average of values all $\le M$. An average of quantities bounded above by $M$ can equal $M$ only if every one of those quantities equals $M$, so $u \equiv M$ on the whole circle $\partial B_r(x_1)$, and by applying the same argument to every radius up to $r$, on the whole disk $B_r(x_1)$. Hence a neighbourhood of $x_1$ lies in $S$, so $S$ is open.

Since $\Omega$ is connected and $S$ is a nonempty subset that is both open and closed in $\Omega$, we must have $S = \Omega$, i.e. $u \equiv M$ throughout $\Omega$. But $u$ was assumed non-constant, a contradiction. Therefore no interior maximum exists; the same argument applied to $-u$ (which is also harmonic) rules out an interior minimum, so both extremes occur only on $\partial\Omega$.