Let $u$ be a non constant real-valued harmonic function in $\mathbb{C}$ Prove that the set $u^{-1}(c)$ is unbounded for every real number $\mathbb{C}$.
This problem is there in the book "Complex Function Theory, Sarason". I found the answer here
The whole proof I got. but the only thing I didn't understood is : Due to continuity of $u$ and connectedness either $u(z)\ge c$ or $u(z)\le c$.
If $u^{-1}(c)$ is bounded then $u(z) \ne c$ outside of some disk with radius $R$. Then $$ \{ z : |z| > R \} = \{ z : |z| > R, u(z)<c \} \cup \{ z : |z| > R, u(z)>c \} $$ is a decomposition of a connected set into two disjoint open sets, so one of the sets on the right-hand side must be empty. (This is essentially a repetition of the argument that a continuous function maps connected sets to connected sets.)
W.l.o.g assume that $u(z)>c$ for $|z| > R$, otherwise replace $u$ by $\tilde u = 2c - u$.
Also $u$ is bounded below on the compact set $|z| \le R$, and therefore bounded below on all of $\Bbb C$, i.e. $u + M$ is a positive harmonic function for some constant $M$.
Then continue as in Does there exist a harmonic function in the whold plane that is postive everywhere?.
