Unless I'm missing something, don't you get a counterexample if $N=M[s]$ where $s$ is a Sacks real over $M$? The point is that every dense open subset of $2^{<\omega}$ in a Sacks extension includes such a dense open set in the ground model and so a Cohen real over $M$ remains Cohen over $M[s]$.
This is a standard fusion argument. One can proceed as follows:
Lemma:
Suppose $p\Vdash``\text{$\dot D$ is dense open in $2^{<\omega}$}"$, $n<\omega$, and $\sigma\in 2^{<\omega}$. Then there are $q\leq p$ and $\tau\in 2^{<\omega}$ such that
$\tau\supseteq \sigma$,
$q\upharpoonright\vphantom{B}^n2= p \upharpoonright\vphantom{B}^{n}2$, and
$q\Vdash \check{\tau}\in\dot D$
proof: Let $\{a_i:i<k\}$ list $p\cap\vphantom{B}^n2$, and for each $i<k$ let $p_i$ be those elements of $p$ that extend $a_i$ (so $p_i\in P$). We define objects $q_i$ and $s_i$ by recursion for $i<k$ so that
- $q_i$ is an extension of $p_i$,
- $s_i\in 2^{<\omega}$,
- $s_0\supseteq\sigma$,
- for $i<k$ we have $s_{i+1}\supseteq s_i$, and
- $q_i\Vdash``\check{s}_i\in \dot D".$
We can do this easily because each $p_i$ is an extension of $p$ in $P$, and hence $p_i$ forces $\dot D$ to be dense open. In fact, for ANY $s\in 2^{<\omega}$ and condition $r\leq p$, there are $r'\leq r$ in $P$ and $t\in 2^{<\omega}$ such that $t\supseteq s$ and $r'\Vdash ``t\in \dot D"$. (We are asking $r'$ to decide a particular value for some extension of $s$ in $\dot D$, and such conditions are dense below $p$.)
Let $\tau= s_{k-1}$, so $\sigma\subseteq s_0\subseteq\dots\subseteq s_{k-1}=\tau$, and since $p$ forces $\dot D$ to be open, it follows that each $q_i$ forces $\tau$ to be in $\dot D$.
Now paste the conditions $q_i$ back together to build $q$. We have
- $q\leq p$ in $P$, and
- $q\upharpoonright\vphantom{B}^n2= p \upharpoonright\vphantom{B}^{n}2$.
Since any extension of $q$ is compatible with some $q_i$, it follows that
- $q\Vdash \check{\tau}\in\dot D$,
and since $\tau$ extends $\sigma$, we are done. $\square$
We use the Lemma to build a fusion sequence. I haven't introduced a definition for $\leq_n$ on $P$, but the above Lemma is strong enough to get what we want for any of the standard variants.
Let $\langle \sigma_n:n<\omega\rangle$ enumerate $2^{<\omega}$, and suppose $p\Vdash``\dot{D}\text{ is dense open in }2^{<\omega}\text{''}$. We build sequences $\langle p_n:n<\omega\rangle$ and $\langle \tau_n:n<\omega\rangle$ such that
- $p_0=p$,
- $p_{n+1}\leq_n p_n$, and
- $p_{n+1}$ forces that $\tau_n$ is an extension of $\sigma_n$ in $\dot D$
Let $q\leq p_n$ for all $n$ (by fusion) and let $E$ be the those elements of $2^{<\omega}$ which extend $\tau_n$ for some $n$. Then $E$ is a dense open subset of $2^{<\omega}$ in the ground model, and
$q\Vdash\check{E}\subseteq \dot D,$
as required.
