Under $MA+\neg CH$ there exists a $Q$-set. A $Q$-set is an uncountable set of reals such that in the subspace topology all of its subsets are $F_\sigma$.
Apparently, buried among more general statements and longer proofs somewhere in this paper it is proven that

If Martin's Axiom holds then every set of reals of cardinality greater than $\omega$ and less than $2^\omega$ is a $Q$-set.

All of the results in that paper seem much more general than I'll ever need and anyway I haven't been able to find any that seem (to me) relevant to the present problem.
I'm wondering if anybody has a reference to where I can find this result or knows how to prove it. I actually only need the existence of a $Q$-set, not the stronger result stated here. 
Thank you.
 A: This is theorem 12 in Martin's Axiom by M.E. Rudin.
I transcribe it here with details filled in and typos fixed:

Lemma: (MA) Suppose $\mathscr{A}$ and $\mathscr{B}$ are families of cardinality $<2^\omega$ of subsets of $\omega$ such that for all finite $\mathscr{C}\subseteq\mathscr{A}$ and all $B\in\mathscr{B}$ the difference $B\smallsetminus\bigcup\mathscr{C}$ is infinite. Then there is an $M\subseteq\omega$ such that $A\cap M$ is finite for all $A\in\mathscr{A}$ and $B\cap M$ is infinite for all $B\in\mathscr{B}$.

Index $\mathscr{A}=\{A_\alpha\}_{\alpha\in\lambda}$ and $\mathscr{B}=\{B_\alpha\}_{\alpha\in\lambda}$ with $\lambda<2^\omega.$
Let
$$\mathbb{P}=\{(k,H)\mid k\text{ is a finite subset of }\omega,\: H\text{ is a finite subset of }\lambda\}$$
And $(k,H)≥(k',H')$ if $k\subseteq k'$, $H\subseteq H'$, and $(k'\smallsetminus k)\cap (\bigcup_{\alpha\in H}A_\alpha)=\emptyset.$
Any uncountable subset of $\mathbb{P}$ has two elements with the same first coordinate. As $(k,F\cup H)$ is below both $(k,H)$ and $(k,F)$ we conclude $\mathbb{P}$ is ccc.
For each $\alpha<\lambda$ and $n\in\omega$ define
$$D_\alpha = \{(k,H)\mid \alpha\in H\}$$
$$E_{\alpha,n}=\{(k,H)\: : \: |k\cap B_\alpha| > n\}$$
Fix $(k,H)$. The $D_\alpha$ are dense as $(k,H)≥(k,H\cup\{\alpha\})\in D_\alpha$. And the $E_{\alpha,n}$ as the difference $B_\alpha\smallsetminus \bigcup_{\xi\in H}A_\xi$ is infinite and therefore $n$ elements may be taken from it and adjoined to $k$.
The family $\mathscr{D}$ is of cardinality $\lambda<2^\omega$ and hence by MA there is a compatible set $Q$ that meets every element of $\mathscr{D}$. Define $L=\bigcup\{k\mid (k,H)\in Q\}$.
Let $A_\alpha\in\mathscr{A}$ then as $Q$ meets $D_\alpha$ there is some $(k,H)\in Q$ with $\alpha\in H$. Now if $a\in A_\alpha \cap L$ then $a\in (k',H')$ for some $(k',H')\in Q$, let $(f,J)$ be under $(k,H)$ and $(k',H')$. Then $a\in f$, but $a\in H$ implies $a\notin f\smallsetminus k$, therefore $a\in k$. This gives $A_\alpha \cap L\subseteq k$ which is a finite set. 
Let $B_\alpha\in B$, for arbitrary $n$ there is a $(k,H)\in Q$ with $|k\cap B_\alpha|>n$, and therefore the intersection $B_\alpha\cap \bigcup\{k\mid (k,H)\in Q\}$ is infinite.

(MA) Every set of reals $X$ such that $\omega<|X|<2^{\omega}$ is a $Q$-set.

Let $Y\subseteq X$ be arbitrary, and $\{U_n\}_n$ be a countable basis for $\mathbb{R}$. We find a $G_\delta$ set $\mathcal{G}$ such that $Y=X\cap\mathcal{G}$. Index (allowing repetitions) $Y=\{y_\alpha\}_{\alpha\in\lambda}$ and $X\smallsetminus Y =\{x_\alpha\}_{\alpha\in\lambda}$, we may assume neither of these sets is empty. Let
$$A_\alpha=\{n\mid x_\alpha\in U_n\}\text{ and }B_\alpha=\{n\mid y_\alpha\in U_n\}.$$
And let
$$\mathscr{A}=\{A_\alpha\}_{\alpha<\lambda}\text{ and }\mathscr{B}=\{B_\alpha\}_{\alpha<\lambda}.$$
Let $\{A_\alpha\}_{\alpha\in F}\subseteq\mathscr{A}$ be a finite subset of $\mathscr{A}$ and $B_\alpha\in\mathscr{B}$.
There is an open set about $x_\alpha$ disjoint from $\{y_\alpha\}_{\alpha\in F}$, and inside this open set infinitely many $U_n$ such that $x\in U_n$ and $U_n\cap\{y_\alpha\}_{\alpha\in F}=\emptyset$, therefore $\mathscr{B}\smallsetminus \bigcup\{A_\alpha\}_{\alpha\in F}$ is infinite.
By the lemma there is an $L\subseteq\omega$ such that for every $\alpha$ the set $L\cap A_\alpha$ is finite and the set $L\cap B_\alpha$ is infinite. Define $L_n = \bigcup\{U_m\mid m\in L, m≥n\} $. Then $\mathcal{G}=\bigcap_{n\in\omega} L_n$ is the $G_\delta$ we were looking for: every element of $Y$ is in every $L_n$ (and hence in $\mathcal{G}$) and every element of $X\smallsetminus Y$ is only finitely many $L_n$ (and hence not in $\mathcal{G}$).
