Weak maximum principle. Suppose $\max_{\bar\Omega}u = M$ is attained at an interior point $x_0 \in \Omega$. At a maximum: $\nabla^2 u(x_0) \leq 0$ (sum of second derivatives is non-positive). But $\nabla^2 u = 0$, so $\nabla^2 u(x_0) = 0$, which is consistent. A direct argument is needed.

Proof via mean value property. For harmonic $u$: $u(x_0) = \frac{1}{|B_r|}\int_{B_r(x_0)} u\,dx$ (mean value property over balls). If $u(x_0) = M = \max u$, then since $u \leq M$ everywhere: $M = \frac{1}{|B_r|}\int_{B_r} u\,dx \leq M$. Equality implies $u = M$ a.e. on $B_r$. By continuity, $u = M$ on $B_r(x_0)$.

Strong maximum principle. Let $S = \{x \in \Omega : u(x) = M\}$. We showed $S$ is open (every point of $S$ has a ball neighborhood in $S$). $S$ is also closed in $\Omega$ (as the preimage of $\{M\}$ under the continuous function $u$). Since $\Omega$ is connected: $S = \emptyset$ or $S = \Omega$. If $u$ attains its maximum at any interior point, $S \neq \emptyset$, so $S = \Omega$ and $u$ is constant.

Mean value property (proof sketch). For harmonic $u$ and ball $B_r(x_0) \subset \Omega$: define $v(\rho) = \frac{1}{|\partial B_\rho|}\oint_{\partial B_\rho} u\,dS$ (surface average). Differentiate: $v'(\rho) = \frac{1}{|\partial B_\rho|}\oint_{\partial B_\rho}\frac{\partial u}{\partial r}\,dS = \frac{1}{|\partial B_\rho|}\int_{B_\rho}\nabla^2 u\,dx = 0$. So $v(\rho) = v(0) = u(x_0)$, proving the mean value property. $\square$