Suppose $L^\infty(X;\mu)$ is separable, then $\mathcal{F}=\{f\in L^\infty(X): \text{Range}(f)\subseteq\{0,1\}\}$ is separable in $\lVert\cdot\rVert_\infty$ norm.
I have proved that for $f,g\in\mathcal{F}$ with $f\ne g$, we have $\lVert f-g\rVert_\infty=1$. Hence, $\mathcal{F}$ is discrete space. As $\mathcal{F}$ is separable, $\mathcal{F}$ should be countable.
From here, I want to prove $L^\infty(X;\mu)$ is finite dimensional. WLOG assume $\mu(X)>0$. If there is sequence of disjoint sets $A_n$ with positive measure, then I can construct for any $I\subseteq \Bbb{N}$, $f=\sum\limits_{n\in I} \chi_{A_n}$. These form an uncountable collection of functions taking values in $\{0,1\}$, contradiction. So, there won't be such $\{A_n\}$.
If I can write $X=\bigsqcup\limits_{k=1}^n A_k$ with $x_k\in A_k$ such that $\mu(\{x_k\})>0$. Then for any $f\in L^\infty(X;\mu)$, we have $f=\sum\limits_{k=1}^n f(x_k)\chi_{A_k}$. So, $\text{Span}\{\chi_{A_k}: k=1,\ldots,n\}=L^\infty(X;\mu)$. But I cannot prove that $X$ should look like this.
Can anyone help me complete the proof? Thanks for your help in advance.