Producing no non-constructible reals The following is stated without proof in Shelah's book "Cardinal arithmetic" (page 276), and is attributed to Uri Abraham:

Suppose that $L[A], L[B]$ have no non-constructible reals and that $\aleph_1^{L[A,B]}=\aleph_1^L.$ Then $L[A,B]$ has no non-constructible reals.

How can we prove this result.
 A: (I don't know if this is Abraham's reasoning.) Let us write $\omega_1$ for $\omega_1^L$ throughout. Suppose for a contradiction the assumptions hold but that $L[A,B]$ has nonconstructible reals.
1) First note that the answer is easy if both $A,B \subseteq \omega_1$:for then for some $\alpha < \omega_1$ $L_\alpha[A\cap\alpha, B\cap \alpha]$ has a non-constructible real by Lowenheim Skolem.  But the latter structure is in $L$, which is absurd.
2) Otherwise: first generically (possibly class-generically) extend $L[A,B]$ to a GCH model $L[A,B][G]$ followed by a Jensen style coding to code $A,B,G$ by some $H\subseteq \omega_1$: we then have $L[A,B,G][H]\models
$ ``$V=L[H]$''. Neither of these forcings add reals (being $\omega$-distributive). However by our assumptions $L[H]$ has non-constructible reals; but now we are in the situation of part 1) and get a contradiction as before.
A: I'll give a different reasoning.
Start as in Philip's argument - assume that $L[A], L[B]$ contains no non-constructible reals but $x\in \mathcal{P}(\omega)\cap L[A, B],\,x\notin L$. We want to show that this situation can be pushed down to $\omega_1$, namely that there is a set $X$, $x\in L[X]$, $X\subset \omega_1$ and for every $\alpha < \omega_1$, $X\cap \alpha \in L$.
Assume, WLOG, that $A, B$ are disjoint classes of ordinals (i.e. $A$ is a subclass of the odd ordinals and $B$ is a subclass of the even ordinals) and let $C=A\cup B$. 
Lemma: Assume $0^\# \notin L[C]$. Let $x \in \mathcal{P}(\omega)\cap L[C]$. There is $U\in L$, $|U|=\aleph_1^{L[C]}$ such that $x\in L[C\cap U]$.
Proof: Take countable elementary submodel $x\in M\prec H(\chi)^{L[C]}$ for suitable large enough $\chi$. Now, it's clear that for every $a\in M$, $\exists t\in a\cap C \iff \exists t \in a\cap C \cap On^M$. Using Jensen's covering lemma, we can cover $On^M$ with some constructible set $U$ of size $\aleph_1$, and get that for every $a\in M$, $\exists t\in a\cap C \iff \exists t \in a\cap C \cap U$. 
Now, by induction on the complexity of the formulas and the ordinals in $M$ we can see that $\langle M, \epsilon ,C\rangle \models \phi(a)\iff \langle M, \epsilon ,C\cap U\rangle \models \phi(a)$ and in particular since for every element in $M$ has a code (which is the finite tree of ordinals that codes its construction), we can see that every the $\epsilon$-relation between codes of elements of $M$ are absolute between $L[C]$ and $L[C\cap U]$, and therefore $x\in L[C\cap U]$. QED
Now, in our case, $\aleph_1 = \aleph_1^L$ so we can choose $f:\omega_1 \rightarrow U$ a bijection in $L$ we get that $x \in L[f^{-1} (C)]$, but for every $\alpha < \omega_1$, $f^{-1}(C)\cap \alpha = (f^{-1}(A) \cap \alpha) \cup (f^{-1}(B) \cap \alpha) \in L$
Remark: The covering lemma here is essential, since this fact doesn't hold if we replace $L$ with the core model and there is a measurable (choose $A,B$ to be mutually generic Prikry sequences, and use this observation, due to Prikry, to show that in $K[A,B]$ there is a Cohen real over $K$). 
