Countable intersection of open dense subsets of $\mathbb{R}$ This particular question was asked in my analysis assignment and I was unable to solve it. Kindly help.

Let A be a subset of $\mathbb{R}$ satisfying $A=\bigcap_{n\geq 1} V_n$, where for each n $\geq1$, $V_n$ is an open dense subset of $\mathbb{R}$ . Which 1 of following is correct?

A.  A is non -empty
B A is countable
C A is uncountable
D A is dense in $\mathbb{R}$.
I marked B which is wrong . A is correct and I marked it.For C,D i don't know how to attempt them. SO, Kindly tell me how to reason b,c,d,
Waiting for your help.

A space in which the intersection of countably many dense open sets is dense is called a Baire space, and the tool that you want is the Baire category theorem, specifically, the version that says that every complete metric space is a Baire space: $\Bbb R$ is a complete metric space, so it is a Baire space.
This immediately tells you that $A$ is dense in $\Bbb R$ and hence that (D) is true. Of course the empty set is not dense in $\Bbb R$, so (A) is also true. To dispose of (B) and (C), suppose that $A$ is countable, say $A=\{a_n:n\in\Bbb Z^+\}$. For each $n\in\Bbb Z^+$ let $U_n=\Bbb R\setminus\{a_n\}$. Let
$$\mathscr{W}=\{U_n:n\in\Bbb Z^+\}\cup\{V_n:n\in\Bbb Z^+\}\,;$$
each $U_n$ is dense and open in $\Bbb R$, so $\mathscr{W}$ is a countable family of dense open subsets of $\Bbb R$, and it’s easy to check that $\bigcap\mathscr{W}=\varnothing$, contradicting the Baire category theorem. Thus, (B) must be false, and (C) must therefore be true.
