Closure of a subset on the real line.

Question

Let $Cl(A)$ denote the closure of $A$. Let the underlying metric space be $[a, b]$ and the modulus function. Let $A$ be the subset of $[a, b]$ where $A=[a, b]$\ $B$ where B is a collection of countably many points from $[a, b].$ Then will it be the case that $Cl(A)=[a, b]?$

Answer 1

Yes, $A$ is dense. If $x \in [a,b]$ is not in the closure then there is an open interval $I$ containing $x$ which contains no point of $[a,b]\setminus B$ which means that the entire interval $I$ is contained in the countable set $B$. This is a contradiction.