Let $V\subset\mathbb{C}$ be a connected open set and $u$ a real-valued harmonic function on $V$. Suppose that the set $S=\{p\in V \mid \nabla u(p)=0\}$ has a limit point in $V$, then $u$ is constant.
My solution: Let $q \in V$ be a limit point of $S$, we can find $r>0$ such that $\bar D (q,r)\subset V$, this disk is simply connected, so there is a holomorphic function $F$ such that $u$ is its real part, now the set $\{ p \in D(q,r) \mid F'(p)=\nabla u(p)=0 \}$ has a limit point in $V$, namely $q$. This implies that $F'$ is identically zero on $D(q,r)$, so $F$ is constant on $D(q,r)$, so $u$ is also constant on $D(q,r)$. From this point, how can we conclude that $u$ is constant on $V$?
Danke
This is my idea for the rest of the proof: Let $z \in V$ be arbitrary, $V$ is path-connected (it is open and connected), so there is a path $\gamma$ with $\gamma(0)=q$ and $\gamma(1)=z$. Since $\gamma^*$ is compact then $\gamma^*\subset \bigcup_{i=1}^nD(z_i,r_i)$ where $ \bar D(z_i,r_i) \subset V$ , $q \in D(z_1,r_1)$ , $z \in D(z_n,r_n)$ and $D(z_{i},r_{i}) \cap D(z_{i+1},r_{i+1}) \neq \emptyset$ for $i=1,...,n-1$. Now $u$ is constant on $D(z_{i+1},r_{i+1})$ if it is constant on $D(z_i,r_i)$. And this implies that $u(z)=u(q)$ as required.
Solution using complex analysis: take the harmonic conjugate and to form a holmorphic function. This holomorphic function will be constant on a set with a limit point contained in the domain, and so is identically constant.