Show that there exists $\delta\gt0 $ , for any $x\in K$ the interval $(x-\delta,x+\delta)$ lies entirely in one of the sets $U_j$. Let $K \subset \mathbb{R} $ be a compact set, and let $U_1\cdots U_n ,n\in \mathbb{N} $ be finitely many open sets with $K \subset U_1 \cup\cdots \cup U_n$. Show that there exists $\delta\gt0 $ , for any $x\in K$ the interval $(x-\delta,x+\delta)$ lies entirely in one of the sets $U_j$.  
I was wondering if I can do this by contradiction.
If for any $\delta\gt0$, there exists $x_0\in K$ such that the interval $(x-\delta,x+\delta)$ do not lie in any of the sets $U_j$ hence also do not lie in $K$. So $x_0$ is an isolated point. But $E$ is compact (i.e bounded and closed). Hence $x_0$ contradicts that $E$ is closed.
Any wrongness in the prove?

It seems the proof is not correct. I'll consider other ways to figure out. Every hints would be helpful. Thanks and regards.
 A: Hints: each $U_i$ is a union of open intervals. A finite number of these cover $K$. So the proof actually reduces to the case when each $U_i$ is an open interval. Arrange these intervals so that the left end points are in increasing order. Look at the intersection of adjoining intervals. Take $\delta$ smaller than the  lengths of these intersections. 
A: Some comments on your answer
Your sentence: If for any $\delta\gt0$, there exists $x_0\in K$ such that the interval $(x-\delta,x+\delta)$ do not lie in any of the sets $U_j$ hence also do not lie in $K$. So $x_0$ is an isolated point. But $K$ is compact (i.e bounded and closed). Hence $x_0$ contradicts that $K$ is closed.


*

*The $x_0$ should not be denoted like that. It depends on $\delta$. So better to denote it by $x_\delta$.

*So $x_0$ is an isolated point. Why? You should provide a convincing argument.

*Hence $x_0$ contradicts that $K$ is closed. Why? A closed subset may have isolated points.


Proof of the requested result
For each $x\in K$, there exists $\delta_x >0$ such that $(x-\delta_x,x+\delta_x)$ is included in one of the $U_j$. The set of open intervals $S=\{(x-\delta_x/2,x+\delta_x/2) \ ; \ x \in K\}$ is an open cover of $K$. As $K$ is compact, one can extract from $S$ a finite open subcover $\overline S = \{(x_1-\delta_{x_1}/2,x_1+\delta_{x_1}/2), \dots,(x_m-\delta_{x_m}/2,x_m+\delta_{x_m}/2)\}$.
$\delta = \dfrac{1}{2}\min\limits_{1 \le i \le m} \delta_{x_i}$ is satisfying what we’re requested. Indeed for $x \in K$, it exists $i \in \{1, \dots m\}$ and $j \in \{1, \dots n\}$such that $x \in (x_i-\delta_{x_i},x_i+\delta_{x_i}) \subseteq U_j$.
Then 
$$(x-\delta , x+\delta) \subseteq (x_i-\delta_{x_i},x_i+\delta_{x_i}) \subseteq U_j.
$$
