Proof verification about a property of the topological space $[0,1]$ Part 2

Suppose $$A_1,\dots,A_k$$ are connected open subsets of $$[0,1]$$ such that $$[0,1]=\bigcup_{i=1}^k A_i$$ and $$A_i \not\subseteq A_j$$ for each $$i\ne j$$.

By characterization of connected subsets of $$\mathbb{R}$$ I know that each $$A_i$$ is an interval.

I want to show:

There exist $$0=a_0 real numbers such that $$[a_{i-1},a_i]\subseteq A_i$$ for each $$i\leq k$$ (possibly permuting the $$A_i$$'s)

My argument

Permuting the $$A_i$$'s if necessary let's suppose

$$A_1=[0=b_1,c_1), A_2=(b_2,c_2),\dots, A_{k-1}=(b_{k-1},c_{k-1}), A_k=(b_k,c_k=1]$$

with the ordering $$b_i for each $$i=1,\dots,k-1$$.

Note that it is not possible that there are indices $$i\ne j$$ such that $$b_i=b_j$$ otherwise we would have $$A_i\subseteq A_j$$ or vicecersa.

Note also that we have also $$c_i for each $$i=1,\dots,k-1$$ otherwise we would have $$A_{i+1}\subseteq A_i$$.

I want to show that $$b_{i+1} for each $$i=1,\dots,k-1$$

For absurd assume that $$b_{i+1}\ge c_i$$ and take $$x\in [c_i,b_{i+1}]$$. Since $$x \ge c_i$$ it follows that $$x\notin A_j$$ for each $$j=1,\dots,i$$. Since $$x \leq b_{i+1}$$ it follows that $$x \notin A_j$$ for each $$j=i+1,\dots,k$$. This is a contradiction since $$[0,1]=\bigcup_{i=1}^k A_i$$.

Now the proof ends by chosing $$a_i \in (b_{i+1},c_i)$$.

Is my proof correct?

P.S. Yesterday I posted a similar question Proof verification about a property of the topological space $[0,1]$ and it turned out that not only the proof was wrong, but also the statement. Even if now I have more carefully written the proof, my experience says that the errors are always lurking, so I will be happy if you control it. Thank you! :)