Let $(\Omega, \Sigma, \mu)$ be a measure space and let $(\mathbb{R}, \mathcal{B})$ be a measurable space. A function $f:\Omega\rightarrow \mathbb{R}$ is called simple if $$f(\omega)=\sum_{i=1}^{n}c_i\mathcal{X}_{A_i},$$ where $c_i\geq0$, $A_i\in\Sigma$, and $\mathcal{X}_{A_i}$ denotes the indicator function.
Now I have to prove that $f(\omega)$ takes finitely many values. In order to prove it, I do the following:
The set $\{c_1, c_2,...,c_n\}$ has the maximum number of all possible combinations as $$D=\sum_{k=1}^{n} {n\choose k}.$$ So, the function $f(\omega)$ can have maximum $D+1$ different values, which completes the proof.
Please correct me if my proof is wrong and if anyone would like to shed some light on it!
Your reasoning seems to be right:
For each point, it will belong to some of the $A_i$'s and therefore the values $f$ can give you are obtained from the partial sums obtained by adding the elements of the subsets contained in $\{c_i\}_{i = 1}^n$. Which are finite.