Given a function $f:E\subseteq \mathbb{R} \to \mathbb{R}$ and finite (see edit) collection $\{E_k\} \subseteq E$ of disjoint sets and $d(E_n,E_m)>0 \forall n \neq m$, and $f|_{E_k}$ is constant. Then $f|_{\cup E_k}$ is continuous.
This fact is used in the proof of Lusin's theorem, but I can't see why $f$ must be continuous on the whole union. I would like an explanation of this step on the proof (I would be specially happy if presented a $\epsilon-\delta$ proof of this continuity, because this is what I'm more confortable with)
The whole proof can be found here.
Edit: I don't know if this edit is under the rules of math.stackexchange, but since it was pointed out in the answer, I wrote countable collection instead of finite collection, which is what I aimed for explanation on what I did not understood, so I edited the question to finite collection, because I still wnat to see an explanation of it and it doesn't feel right to create a new question about this. If this edit is not under the rules, please forgive me and feel free to unedit.
The result is false as you have stated it.
Let $E_0=\{0\}$, for $n\in\Bbb Z^+$ let $E_n=\left\{\frac1n\right\}$; clearly $d(E_n,E_m)>0$ whenever $n\ne m$. Let $E=\bigcup_{n\ge 0}E_n$, and define
$$f:E\to\Bbb R:x\mapsto\begin{cases} x,&\text{if }x\in E\setminus E_0\\ 1,&\text{if }x=0\,. \end{cases}$$
Clearly $f$ is constant on each $E_n$, but $f$ is not continuous at $0$.