Exercise on relative constructibility Given a set $A$ in Kunen's set theory book (page 143) we can find the following definition of the relative constructible universe $L[A]$.

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*$L[A](0)=\{A\}\cup \operatorname{tr cl}(A)$

*$L[A](\alpha+1)=\mathcal{D}^+(L[A](\alpha))$

*$L[A](\gamma)=\bigcup_{\alpha<\gamma} L[A](\alpha)$ if $\gamma$ is a limit ordinal.

where $trcl(A)$ denotes the transitive closure of $A$ and $\mathcal{D}^+(L[A](\alpha))$ the collection of all sets defined with parameters in $L[A](\alpha)$.
In this regard, it is known that if $A\subset \omega$ then $L[A]$ satisfies $GCH$. So my question is, if $A\subset \omega$ then does $L[A]$ satisfy $V=L$ and $AC$ and therefore $GCH$ is true in $L[A]$?
Every answer or correction will be appreciated.
 A: There are two notions of relative constructibility, $L[A]$ and $L(A)$. What you describe is actually $L(A)$ and not $L[A]$, and the two coincide when $A$ is a subset of $L$ (e.g. a set of ordinals).
In the $L[A]$ construction we add $A$ as a predicate to the language and we construct the model by starting with the empty set, and taking the definable (with parameters) subsets of the structure $(L_\alpha[A],\in,A\cap L_\alpha[A])$.
You can prove that $L[A]$ is always a model of $\sf ZFC$, whereas $L(A)$ is a model of $\sf ZFC$ if and only if the transitive closure of $\{A\}$ has a definable well-ordering (so in most cases, choice fails in $L(A)$).
To make matters worse, a lot of people confuse the two notations often, or simply don't care enough to discern them. But there is a significant difference between $L[\Bbb R]$ and $L(\Bbb R)$ in certain contexts.
But as I said, if $A$ is a set of ordinals, then the two things are the same so there's no room for confusion. Now to your question.
It is certainly not true that $L[A]\models V=L$, this will be true if $A\in L$, but if $A$ is not constructible itself, then this is no longer true. The proof, however, is somewhat similar to the proof that $\sf GCH$ holds in $L$. You can even show that if $A\subseteq\omega_1$ then $\sf GCH$ holds in $L[A]$. 
It need not be true for a subset of $\omega_2$ since we can force over $L$ to add $\aleph_2$ generic reals and code them as a subset $A$ of $\omega_2$, so $L[A]\models 2^{\aleph_0}=\aleph_2$ in that case.
