I have the following theorem and its proof but I can't understand some steps of the proof I hope you can help me.
If X is a chainable continuum and $C = \{U_{1}, . . . , U_{n}\}$ is a ε–chain in $X$ covering it then there is an ε–chain $C''= \{U'_{1}, . . . , U'_{n}\}$ in $X$ which covers it such that for all $k \in \{1, . . . , n − 2\}$, $\overline{U'_{k}} \cap \overline{U'_{k+2}} = \emptyset$.
Proof.
We have $X- \cup_{k=2}^{n} U_{k}$ is closed in X then is compact and $X- \cup_{k=2}^{n} U_{k} \subset U_{1}$.
There is an open $U'_{1}$ such that $X- \cup_{k=2}^{n} U_{k} \subset U'_{1} \subset \overline{U'_{1}} \subset U_{1}$. So $\{U'_{1}, U_{2} . . . , U_{n}\}$ is a ε–chain in $X$. [Why? My doubt here is why $U'_{1} \cap U_{2} \neq \emptyset$].
Following the above procedure we now have $X-\{U'_{1} \cup \cup_{k=3}^{n} U_{k} \}$ is compact in X and $X-\{U'_{1} \cup \cup_{k=3}^{n} U_{k} \} \subset U_{2}$ so there is an open $U'_{2}$ such that $X-\{U'_{1} \cup \cup_{k=3}^{n} U_{k} \} \subset U'_{2} \subset \overline{U'_{2}} \subset U_{2}$ then $\{U'_{1}, U'_{2}, U_{3}, . . . , U_{n}\}$ is a ε–chain in $X$. Continuing in this way we build $C''$.
My other question is why do we have to $\overline{U'_{k}} \cap \overline{U'_{k+2}} = \emptyset$.
A metric space (X,d) is chainable if for every $\epsilon>0$, there is an $\epsilon$-chain covering X. An $\epsilon$-chain is a simple chain $C=\{U_{1},...,U_{n}\}$ of open sets where the diameter of each $U_{i}$ is less than $\epsilon$. And a family $\{U_{1},...,U_{n}\}$ of subsets of a metric space (X,d) is a simple chain in X if we have $U_{j} \cap U_{k} \neq \emptyset$ if and only if $|j-k| \leq 1$.
Any help is appreciated.