If E is a set of positive measure, $\{E_n\}$ is a countably infinite set of pairwise disjoint subsets of E. for each nonempty subset S of positive integers, we let $\mathcal{X}_s ={}$the characteristic function of $\bigcup_{n\in s} E_n$ prove that if $ S \neq T $ then $ \| \mathcal {X}_S - \mathcal {X}_T \|_\infty = 1$
SO
What I know that the characteristic function $\mathcal{X}_s = 1 $ if $n\in s $ and $0$ if $ n\notin s$ and I know that $\| \mathcal{X}_s \|_\infty = \sup_{1\le s \lt \infty} |\mathcal{X}_s|$ which means the essential upper bound we learn that $\|f + g\|_\infty \le \|f\|_\infty + \|g\|_\infty$ But I think this won't help me solve my problem . what I did is when I subtract two characteristic functions I get zero since 1-1 is zero but I didn't get one. I would like to hear your advice how to solve this. Thank in advance.
If $S \neq T$, then (without loss of generality), there is some $k \in S$ such that $k \not\in T$. Then since the sets $\{E_n\}$ are non-empty and disjoint, there is $x \in E_k$ such that $x \not \in E_n$ for any $n \in T$. At this $x$, we have $\mathcal X_S(x) = 1$ and $\mathcal X_T = 0$. Thus $$\lvert \mathcal X_S(x) - \mathcal X_T(x) \rvert = 1$$ and passing to the supremum, you will have $$\|\mathcal X_S - \mathcal X_T \|_\infty \ge \lvert \mathcal X_S(x) - \mathcal X_T(x) \rvert= 1.$$ (It is also easy to show the supremum can be no more than $1$ and thus conclude that $\|\mathcal X_S - \mathcal X_T \|_\infty = 1$).
Note that this is only true if $\| \cdot \|_\infty$ is the honest-to-goodness supremum. If it is the essential supremum (as is used in $L^\infty$), then there are unequal sets $A,B$ such that $\| \chi_A - \chi_B\|_{\infty} =0$. Indeed, $A = (-1,1)$ and $B = (-1,1) \cup \{ 5\}$ will serve as an example. Similarly, in your case, if $E_1$ has measure zero and $E_2$ has positive measure, then you can put $S = \{1,2\}$ and $T = \{2\}$ and you'll have $$\|\mathcal X_S - \mathcal X_T \|_\infty=0.$$ To fix this, you could require that all $E_n$ have positive measure, and then the statement continues to hold when using the essential supremum.