Is it true that every closed set is a countable intersection of open intervals?

Motivation:

We know that every open set is a countable union of open intervals with rational endpoints and that every open interval is a countable union of closed intervals. Hence every open set is a countable union of closed intervals. It follows by De Morgan's laws that every closed set is a countable intersection of open sets.

I would like to ask if we can prove a stronger result that

every closed set is a countable intersection of open intervals.

• This is not a duplicate. – Kavi Rama Murthy Feb 20 at 9:18
• Why is this question closed and receive up to 3 downvotes? – Le Anh Dung Apr 16 at 14:02

For example (as @AlbertoTakase mentions in the comments below), consider the set $$\{0\} \cup \{1\}$$. This is a closed set since finite subsets of $$\Bbb{R}$$ are closed, but it is clearly not a closed interval. Hence, it cannot be written as a countable intersection of open intervals.
• (to add an example) Consider $\{0\}\cup\{1\}$ which is closed but not an interval. – Alberto Takase Feb 20 at 9:15