Definition. A function $s\colon \to\mathbb{R}$ is said $\textit{simple}$ if $s(X)=\{c_1,\dots,c_n\}.$
Proposition. Let $s\colon X\to\mathbb{R}$ a simple function. Then they exist and are unique $c_1,\dots,c_n\in\mathbb{R}$ distinct and $E_1,\dots ,E_n\subseteq X$ disjoint such that $$X=\bigcup_{k=1}^n E_k$$ and $$s=\sum_{k=1}^n c_k\chi_{E_k}.$$
Proof. For all $k\in\{1,\dots, n\}$ we place $E_k:=\{s=c_k\}.$ It's clear that $E_i\cap E_k=\emptyset$ for $i\ne j$ and $X=\cup_{k=1}^n E_k$, moreover $$s=\sum_{k=1}^n c_k\chi_{E_k}.$$ We show the uniqueness. Let $d_1,\dots,d_m$ distinct and $F_1,\dots, F_m\subseteq X$ disjoint, with $$X=\bigcup_{l=1}^m F_l$$ such that $$s=\sum_{l=1}^{m} d_l\chi_{F_l}.$$ As $$\sum_{k=1}^n\chi_{E_k}=1,\quad\sum_{l=1}^m\chi_{F_l}=1$$ we have that $$s=\sum_{k=1}^n\sum_{l=1}^mc_k\chi_{E_k}\chi_{F_l}=\sum_{k=1}^n\sum_{l=1}^md_k\chi_{E_k}\chi_{F_l},$$ therefore $$\sum_{k=1}^n\sum_{l=1}^m(c_k-d_k)\chi_{E_k\cap F_l}=0\tag1.$$ As the sets $\{E_k\cap F_l\}_{k\in\{1,\dots,n\},l\in\{1,\dots, m\}}$ are disjointed and as $$X=\bigcup_{k=1}^n\bigcup_{l=1}^m E_k\cap F_l,$$ for all $k\in\{1,\dots, n\}$ exists a unique $l_k\in\{1,\dots, m\}$ such that $c_k=d_{l_k}$ and for all $l\in\{1,\dots,m\}$ exists a unique $k_l\in\{1,\dots, n\}$ such that $c_{k_l}=d_l$. Then $n=m$ and for all $k\in\{1,\dots, n\}$ we have that $c_k=d_{l_k}$ and $E_k=\{s=c_k\}=\{s=d_{l_k}\}=F_{l_k}$
Question For all $k\in\{1,\dots, n\}$ exists a unique $l_k\in\{1,\dots, m\}$ such that $c_k=d_{l_k}$ and for all $l\in\{1,\dots,m\}$ exists a unique $k_l\in\{1,\dots, n\}$ such that $c_{k_l}=d_l$. Then $n=m$ and for all $k\in\{1,\dots, n\}$ we have that $c_k=d_{l_k}$ and $E_k=\{s=c_k\}=\{s=d_{l_k}\}=F_{l_k}.$
Why can we say this?
Thanks!
After hint by gandalf61, this is my answer.
From $(1)$ we have that for all $k\in\{1,\dots, n\}$ exists unique $l_k\in\{1,\dots, m\}$ such that $E_k\cap F_{l_k}=\emptyset$, otherwise the hypothesis according to which the $d_1,\dots, d_m$ are distinct would be violated. We consider the map $f\colon\{1,\dots,n\}\to\{1,\dots, m\}$ defined as $k\mapsto f(k):=l_k$. The map $f$ is well defined, moreover $f$ is injective, indeed fixed $k\ne j$ in $\{1,\dots, n\}$ they are unique $l_k, l_j\in\{1,\dots, m\}$ such that $c_k=d_{l_k}$ and $c_j=d_{l_j}$. Since $c_1,\dots, c_n$ are distinct $c_k\ne c_j$ then $d_{l_k}\ne d_{l_j}$ therefore $l_k\ne l_j$. In the same way it is shown that the map $g\colon\{1,\dots,m\}\to\{1,\dots, n\}$ defined as $l\mapsto g(l):=k_l$ is injective. Therefore $n=m$. To recap: $n=m$ and for all $k\in\{1,\dots, n\}$ exists unique $l_k\in\{1,\dots, n\}$ such that $c_k=d_{l_k}$, then for all $k\in\{1,\dots, n\}$ $$E_k=\{s=c_k\}=\{s=d_{l_k}\}=F_{l_k}.$$