A continuous plane function with the local mean-value property is harmonic

Fix a closed disc $\overline{D(a,R)}\subseteq \mathrm{\Omega }$. By [L1], the boundary values $t\mapsto u(a+R{e}^{it})$ have a unique continuous harmonic Poisson extension $v$ to $\overline{D(a,R)}$, and [L2] makes $v$ satisfy the same mean-value property there. Put $w:=u-v$. Then $w$ is continuous on $\overline{D(a,R)}$, has the local mean-value property on $D(a,R)$, and vanishes on the boundary circle.

By [L3], the closed disc is compact, so $w$ attains a maximum $M$ and a minimum $m$ on $\overline{D(a,R)}$. If $M>0$, then the boundary values being $0$ force the maximum to occur at some interior point $b$. For a small circle centered at $b$, the circle mean-value property gives $M=w(b)$ as the average of values all bounded above by $M$, so every value on that circle is also $M$. Repeating this argument on overlapping small circles shows that the set $\{w=M\}$ is both open and closed in the connected disc, hence all of $D(a,R)$; this contradicts the boundary value $0$. Therefore $M\le 0$.

Applying the same argument to $-w$ gives ${\max }_{\overline{D(a,R)}}(-w)\le 0$, so $-w\le 0$ and therefore $w\ge 0$ on the closed disc. Together with step 2.1 this gives $w=0$ on $\overline{D(a,R)}$, so $u=v$ there.

Since the closed disc was arbitrary and $v$ is harmonic on its interior, $u$ is harmonic on every point of $\mathrm{\Omega }$, hence on all of $\mathrm{\Omega }$.