Let $f: \Omega \rightarrow \mathbb{C}$ holomorphic such that $\Omega \subseteq \mathbb{C}$. Let $z_0 \in \Omega$ such that $\overline{D_{R}}\left(z_{0}\right) \subseteq \Omega$. Given that for every $z \in \partial D_{R}\left(z_{0}\right)$ $f(z) \in \mathbb{R}$, prove that $$f(z_0)\in \mathbb{R}$$
so I was trying to use cauchy integral theorem : $$f(z_0) = \frac{1}{2\pi i} \int_{\partial D_{R}(z_0)} \frac{f(z)}{z- z_0}dz$$ so I know $f(z) \in \mathbb{R}$ but I am not sure how should I procced from here
parameterize your curve (the boundary of the disk) by $\partial D_R=\{z_0+Re^{it}:0\leq t\leq 2\pi\}$. With this, your integrand becomes $$\frac{f(z)}{z-z_0}=\frac{f(z)}{Re^{it}}$$ and your measure for integration will be $$dz=iRe^{it}dt$$ So the integral will become $$\frac{1}{2\pi i}\int_0^{2\pi}\frac{f(z)}{Re^{it}}iRe^{it}dt=\frac{1}{2\pi }\int_0^{2\pi}f(z)dt\in \mathbb{R}$$
The point to be taken away from this is that for holomorphic functions, their value at a point is dictated by their values on the boundary of a disk containing that point throughout which the function is holomorphic. This is exactly what the Cauchy integral formula says, and this further drives that point home.