Consider $I=[0,1]$. Assume a priori we are given a collection of open sets $(U_{t})_{t\in [0,1]}$, $U_t$ open in $I$ such that $I=\bigcup_{t\in I}U_t$. Here, $U_t\owns t$.
Problem is: Show that there exists $0=t_0<t_1<....<t_k=1$ such that $U_{t_i}\cap U_{t_{i+1}}\neq \varnothing$ for all $i$ and $I=\bigcup_{i=1}^kU_{t_i}$
My attempt:
Since $I$ is compact, find $w_1,w_2,...,w_n$ such that $I=\bigcup_{i=1}^nU_{w_i}$. Adjoin $U_0$ and $U_1$ if necessary so we may assume $w_0=0$ and $w_n=1$.
Set $t_0=w_0=0$. We must find $1>t_1>t_0$ such that $U_{t_1}\cap U_{t_0}\neq\varnothing$.Since $U_{t_0}$ is open, there exists an $r>0$ such that $B_{r}(t_0)\subseteq U_{t_0}$. Fix any $t_1\in B_{r}(t_0)$ such that $1>t_1>t_0=0$. Then, $U_{t_1}\cap U_{t_0}\neq \varnothing$.
Can I ensure that there exists $1>t_2>t_1>t_0=0$ such that $U_{t_2}\cap U_{t_1}\neq \varnothing$. If it is so, then one can do induction.
I think one should implement the pigeonhole principle. Is there an easier way?