Let $(a,b)\in \mathbb R^n$. We consider the following open cover of the compact line segment $[a,b]$: $$[a,b]\subset\underset{x\in [a,b]}{\bigcup}B(x,\rho\_x),$$
where for $x\in K,B(x,\rho\_x)$ is a given nonempty open ball of center $x$.


Since $[a,b]$ is compact, there exist a finite subcover such that
$$[a,b]\subset\underset{1\leq k\leq n}{\bigcup}B(x\_k,\rho\_{x\_k}).$$


I would want to know if we can always construct the above subcover such that 
$$x\_{k+1}\in B(x\_k,\rho\_{x\_k}),$$
for all $k\in\{1,\ldots,n-1\}$.



Special finite subcover of a compact
The answer is no. Consider the interval $[0,2]$, and define $\rho\_x$ as follows:


1- For $x=1$, let $\rho\_x=0.5$


2- For $x\neq 1$, Let $\rho\_x$ be the distance from $x$ to $1$, so $\rho\_x=|1-x|$


Now suppose that a subcovering $B(x\_i,\rho\_i)$, $i=1\cdots,n$ of the wanted form exists. By the choice of the $\rho\_x$, the unique open ball $B(x,\rho\_x)$ containing $1$ is $B(1,\rho\_1)$. Thus $B(1,\rho\_1)$ must be the first ball of the subcovering, and without loss of generality, we can suppose that $x\_2<1$. Then by the choice of $\rho\_2$, $B(x\_2,\rho\_2)\subseteq (0,1)$. Since $x\_3\in B(x\_2,\rho\_2)$, then $x\_3<1$ and $B(x\_3,\rho\_3)\subseteq (0,1)$ for the same reason as above. 


So by an easy induction, we show for all $i=2\cdots,n$ that $B(x\_i,\rho\_i)\subset (0,1)$. The union of the $B(x\_i,\rho\_i)$ for $i=1,\cdots,n$ is then contained in $[0,\frac{3}{2})$. Contradiction.


