let $\{V\_n\}\_{n=1}^{\infty}$ be sequence of open and dense subsets of $\mathbb{R}^{\mathbb{N}}$. set $A= \cap\_{n=1}^{\infty}V\_n$
then A is

choose the correct option

$1.)$ open

$2.)$ closed

$3.)$ countable

$4.)$ uncountable

My attempt : I take $n = 1.$

Since $\mathbb{R}$ is a complete metric space, the Baire category theorem implies that $A$ is dense, in particular it is nonempty.
We also claim that $A$ has to be uncountable. Indeed, suppose that $A=\{x\_1, x\_2, \ldots \}$ was countable, and define the sets $W\_n := \mathbb{R} \smallsetminus \{x\_n\}$. Then clearly each $W\_n$ is an open dense subset of $\mathbb{R}$, and thus applying the Baire category theorem to the countably many sets $V\_n,W\_n$ yields that $\bigcap\_{n \geq 1} (V\_n \cap W\_n)$ is dense in $\mathbb{R}$, in particular it is nonempty. On the other hand
$$ \bigcap\_{n \geq 1} (V\_n \cap W\_n) = \left(\bigcap\_{n \geq 1} V\_n \right) \cap \left( \bigcap\_{n \geq 1} W\_n \right) = A \cap (\mathbb{R} \smallsetminus \cup\_{n \geq 1} \{ x\_n \} ) = A \cap (\mathbb{R} \smallsetminus A) = \emptyset.$$
This is a contradiction, so $A$ is uncountable. 

Im confusing that A must be open?

Any hints/solution 

thanks u