Let $r_1,r_2,....$ be the list of all rational numbers in $[0,1]$. Let $U_n=\left(r_n-\dfrac1{3^n},r_n+\dfrac1{3^n}\right)$ then $[0,1]\not \subseteq \displaystyle\bigcup_n^\infty U_n$
I cannot prove this because first I dont know we are doing this orderly, I mean If I knew that we start from $0$ and go with this given $U_n$ then I can say that there will be some gap to the $1$ but $r_n$ can be randomly located so their $\left(r_n-\dfrac1{3^n},r_n+\dfrac1{3^n}\right)$ radius can be big to cover whole $[0,1]$ How can I start
$[0,1]$ is compact so $[0,1]] \subset \cup_{n=1}^{N} U_n$ fro some $N$. Now you can arrange the finite number of intervals $U_1,U_2,\cdots,U_N$ with increasing order of left end points and get a contradiction (since the total length of $U_i$'s is less than $1$).