It is well known that for every subset $A\subset \omega_1$ if $V=L[A]$ then $L[A]\vDash GCH$. In particular $L\vDash \exists A\subset \omega_1\,(L[A]\vDash\, GCH)$. Nonetheless, it is also consistent with ZFC the negation of that statement. Specifically, it can be proved that there exists a model of $ZFC$ plus "$\exists A\subset \omega_1\,(L[A]\vDash\neg CH)$". My question is indeed related with this fact.
It's obviuous that the model that we want construct can't be $L$ or whatever which satisfies $V=L[A]$ for some $A\subset \omega_1$. So we would start with a countable transitive model $M$ of $V=L$ and force with the Cohen forcing $Fn((\aleph_2)^M\times\omega, 2)$ to get a generic extension $M[G]$ in which $(V\neq L\;\wedge\; 2^{\aleph_0}=\aleph_2)^{M[G]}$. Now, working in $M[G]$, there exists $A_0\subset \omega_2$ coding a well-order of $\omega_2$ of ordertype $2^{\aleph_0}$ so if we attach $A_0$ to $L$ to bild up $L[A]$ we finally have $M[G]\vDash \exists A\subset \omega_2\,(L[A]\vDash 2^{\aleph_0}=\aleph_2)$.
Here is the point where my question arise. If now, we would force over $N=M[G]$ with the forcing $Fn(\omega,(\omega_1)^{M[G]})$ we would construct a generic extension where $\omega_2^{N}=\omega_1^{N[H]}$ so that $A_0$ is a subset (in $N[H]$) of $\omega_1$. However I'm not sure if $(L[A]\vDash 2^{\aleph_0}=\aleph_2)^{N[H]}$ is the case because I have not any information about $(2^{\aleph_0})^{N[H]}$.
So, could somebody help me to prove that in the generic extension $N[G]$ the statement $(L[A]\vDash 2^{\aleph_0}\neq\aleph_1)^{N[H]}$ holds?
Thank you so much.