Let $ f $ be a non-constant integer function. Given a number $ \rho> 0 $, let's define: $ E_{\rho} = \{z \in \mathbb {C}: |f(z)| <\rho \} $, $ F_{\rho} = \{z \in \mathbb {C}: |f(x)| \leq \rho \} $.
(a) Prove that the adhesion of $ E_{\rho} $ is equal to $ F_{\rho} $.
(b) Justify that in each bounded related component of $ E_{\rho} $ there is at least a zero of $ f $.
My attempt: For the first part, let $ z \in \overline {E _{\rho}}$. Then for all $ \epsilon > 0$ we have that $ B(z, \epsilon) \cap E_{\rho} \neq \emptyset $. Then, we have that $ |f (z)| \leq \rho $, that is, $ z \in F_{\rho} $. I am not totally sure of this. And for the other part I have not had results yet.
I suppose by 'integer function' you mean an analytic function any by 'adhension' you mean closure.
If $|f(z)|=\rho$ then there must be a sequence $z_n \to z$ such that $|f(z_n)| <\rho$ for all $n$: otherwise $|f| \geq \rho$ in some disc around $z$ and this contradicts Maximum Modulus Principle applied to $\frac 1 f$ since $f$ is non-constant. Hence the closure of $E_{\rho}$ is $F_{\rho}$.
The second part also follows from Maximum Modulus Principle: The boundary of any component of $E_{\rho}$ is contained in $\{z: |f(z)|=\rho\}$ (by first part). Suppose there is no zero of $f$ in one such component. We can now apply Maximum Modulus Principle to $f$ and $\frac 1 f$ to see that $f$ must be a constant.