let:$\text{U}\subseteq \mathbb{C}$ as $U$ open and path connected. $f$ analytic in $U$ and does: $\forall z_0\in U,\forall \epsilon >0,\exists z_1,z_2\in Ball_{\epsilon }\left(z_0\right)st:f\left(z_1\right)=f\left(z_2\right)$, prove $f$ is constant.
My try:
deduce in negative that $f$ is not constant, thus $z_1 \ne z_2$, and we will take the domain $Ball_\frac {\epsilon_0} 2(z_0)$.
$f(z)$ is analytic at that ball and at the closure of that ball, so according to the maximum principle, $f$ receives maximum at the boundary, which means:
$Max\left\{\left|f\left(z\right)\right|\right\}=f\left(z_p\right),\:as\:z\in \partial \left(Ball_{\frac{\epsilon }{2}}\left(z_0\right)\right)$, when $z_p$ is the maximum point.
without loss of generosity, $z_2>z_1$ and decide $z_2=z_p$, thus $z_p$ is maximum and it makes $f(z_p)=f(z_2)>f(z_1)$, in contradiction to the suggest that $f(z_2)=f(z_1)$
and then from the identity theorem, just "move" it up to $Ball_\epsilon(z_0)$
Now, I know this solution may have flaws, I am not sure of it since I assumed stuff that maybe are wrong ( probably using the exist and choosing if it is not allowed ) .
If that proof is no good and I cant make it to work using that way, I will just try to understand my professor Answer, he has 2 solutions which are not understandable, but I will try my best to understand if needed