I read this statement when reading a proof about "continuous function with compact support is dense in $L^p$". For simplicity, I consider the case $n=1$, but how do I construct a closed covering $\{U\}_{i=1}^\infty$ that make the a open interval $U=\cup_{i=1}^\infty U_i$. The boundary of the open set can not be covered by closed set and still in $U$, isn't it? For example, the interval $(0,1)$, I can't cover its boundary by closed sets and make the closed sets in it, or should I consider closed sets like $[\frac{1}{n},1-\frac{1}{n}]$?
Another question is that, I also read another statement:
every open set in $\mathbb{R}$ is a countable union of disjoint half-open intervals.
How do I construct this?
In $\mathbb{R}^n$, if a set $U$ is open then for every $x\in U$ you can find an open ball $B_\delta(x)$ of appropriately small radius $\delta$ centered on $x$ entirely contained inside $U$. Thus, the closure of the ball $B_{\delta/2}(x)$ is also contained inside $U$. If you take a union over all of these balls, then you recover $U$.
If you wish to only consider countable unions of closed sets, then we can do something similar but only use balls centered around "completely rational points" with rational radii. The explicit construction is harder to write down, but the idea is the same.