Miller real is not in the closure of sets under some conditions Background
We can define Miller Forcing as the poset of nonempty perfect rational trees. That is, we define:


*

*$p\subset 2^{<\omega}$ is a perfect tree iff it is closed downwards (for all $s, n$, if $s \in p$ and  $n \in \omega$, then $s|n \in p$) and every branch splits (for all $s \in p$ there exists $t \in p$ such that $s\subset t$ and $t^\frown(0), t^\frown(1) \in p$.

*$\mathbb Q\subset 2^{\omega}$ is the set of binary sequences that are eventually $0$ ($\mathbb Q=\{s^\frown \textbf{0}:s \in 2^{<\omega}\}$, where $\textbf{0}$ is the constant zero function.

*Given a perfect tree $p$, $[p]=\{x \in 2^\omega: \forall n \in \omega\,(x|n \in p)\}$. It's possible to show that $[p]$ is a perfect set, and that every perfect set is of this kind.

*A perfect tree $P$ is a rational perfect tree iff $\mathbb Q\cap [p]$ is dense in $[p]$.

*$\mathbb M$ is the set of all perfect rational trees ordered by inclusion.

*$P\subset 2^\omega$ is a rational perfect set if it's nonempty, perfect (closed without isolated points) and $P\cap \mathbb Q$ is dense in $P$.
It's possible to show that being a perfect tree is absolute for transitive models of ZFC, therefore if $M$ is such a model model, then $\mathbb M^M=\mathbb M\cap M$.
Let $M$ be a ctm and $G$ be $\mathbb M^M$-generic over $M$. The Miller real is defined as:
$$f=\bigcup\bigcap G$$
It's possible to show that the domain of $f$ is $\omega$, that $\bigcap G=\{f|n: n \in \omega\}$ and that $G=\{p \in \mathbb M^M: \forall n \in \omega\,(f|n \in p)\}$.
Question
I have read in this article (theorem 4, first implication) that if $F\subset 2^\omega$ and $F \in M$ is such that $(F$ is closed in $2^\omega$ and  $F$ does not contain any perfect rational subset$)^M$, then if $f$ is a miller real, it follows that $f \notin \operatorname{cl} F$. But I don't know why.
I managed to prove that under these hypothesis, $(\operatorname{cl} (F\cap \mathbb Q)$ is countable$)^M$ and therefore $\operatorname{cl}^M (F\cap \mathbb Q)=\operatorname{cl} (F\cap \mathbb Q)$, but I don't know if this helps.
 A: For any closed set $F\in V$, one can show $D_F=\{p\in \mathbb{P}: \exists \text{open }U \ [p]\subset U\& U\cap F=\emptyset\}$ is dense. Just a comment though, I believe when people are talking about Miller forcing, another form (essentially the same) is more common, i.e subtrees of $\omega^{<\omega}$ such that each node has an extension that splits infinitely (has infinitely many immediate extensions). You are welcome to provide more details if you'd like :-). 
A: Let $D=\{p \in\mathbb M^M: \exists s \in p(\forall t \in p(p\subset t \vee t \subset p) \wedge \forall x \in F(s\not\subset x)\}$ (I believe this is the same set as in Jing Zhang's answer). By absoluteness, $D \in M$. We will see that $D$ is a dense subset of $\mathbb M^M$.
Let's work inside of $M$. Suppose $p \in \mathbb M$. There exists $s \in p$ such that for all $x \in F$, $s \not \subset x$. If not, for each $s \in p$ let $x_s \in F$ be such that $s\subset x_s$. Then $\{x_s: s \in p\}\subset F$, and, since $F$ is closed, $\operatorname{cl}\{x_s: s \in p\}\subset F$. But $[p]\subset \operatorname{cl}\{x_s: s \in p\}$, since given $y \in [p]$ and $n \in \omega$, $y|n \in p$, and, therefore, $y|n=x_{y|n}|n$. We have shown that $[p]\subset F$, but $F$ contains no perfect rational subsets, which is a contradiction. So there exists $s \in p$ such that for all $x \in F$, $ s\not \subset x$. Let $p|s=\{t \in p: t\subset s \vee s \subset t\}$. $p|s$ is a perfect rational tree, $p|s\leq p$ and $p|s \in D$. Therefore, $D$ is dense.
Now let's step out of $M$. Let $p \in G\cap D$. Since $p \in G$, $\{f|n: n \in \omega\}\subset p$. Since $p \in D$, there exists $s$ such that for all $s \in p$, $s\subset t\vee t\subset s$ and $\forall x \in F$, $s\not \subset x$. So there exists $n \in \omega$ such that $f|n=s$. If $f \in \operatorname{cl} F$, there must be $x \in F$ such that $f|n=s\subset x$, which is a contradiction.
