Is every first-countable Lindelof space of cardinality $<\mathfrak c$ a $Q$-space under MA? Definition. A topological space $X$ is a $Q$-space if every subset of $X$ is of type $G_\delta$.
It is clear that every $Q$-space has countable pseudocharacter (= all singletons are $G_\delta$) and is perfect (= all closed subsets are $G_\delta$).
It is well-known that Martin's Axiom (MA) implies that every second-countable space of cardinality $<\mathfrak c$ is a $Q$-space.
A standard proof of this fact actually shows more, namely, that every topological space $X$ with countable separating weight and $|X|<\mathfrak c$ is a $Q$-space.
Let us recall that a topological space $X$ has countable separating weight if it admits a countable open cover $\mathcal U$ such that for any distinct points $x,y\in X$ there exists a set $U\in\mathcal U$ that contains $x$ but not $y$.
It is easy to see that every topological space with countable separating weight has countable pseudocharacter. The space $\omega_1$ of ordinals with the order topology is a first-countable but not a $Q$-space.
Question. Let $X$ be a first-countable Lindelof Hausdorff space of cardinality $|X|<\mathfrak c$. Is $X$ a $Q$-space under MA? Is $X$ perfect under MA?
Remark. The "Lindelof" requirement can not be removed from this question as the ordinal $\omega_1=[0,\omega_1)$ with the standard order topology is first-countable but not perfect (because the closed set of countable limit ordinals is not of type $G_\delta$ in $\omega_1$ by Fodor's Lemma) and hence $\omega_1$ is not a $Q$-space.
 A: I think that the answer to both of your questions is negative.
Let $X$ be a subspace of the real line with its usual topology of size $\omega_1$. Then, clearly, $X$ is first countable, Lindelöf and Hausdorff. Thus, its Alexandroff duplicate, $A(X)$, is also  first countable, Lindelöf and Hausdorff (see here The Alexandroff Duplicate and its subspaces ). Now, since every point in $X\times \{1\}$ is isolated, it is clear that $X\times \{1\}$ is a uncountable and discrete subspace of $A(X)$; in particular, $X\times \{1\}$ is not Lindelöf. Since every Lindelöf perfect space is hereditarily Lindelöf, it follows that $A(X)$ is not perfect.
Hence, $A(X)$ is a first countable, Lindelöf and Hausdorff space of cardinality $\omega_1$ which is not perfect (thus, not a $Q$-space). Finally, since $\mathsf{MA}+\neg \mathsf{CH}$ is consistent, this procedure would in turn give an example of a first countable, Lindelöf and Hausdorff space of cardinality $<\mathfrak{c}$ which is not perfect (thus, not a $Q$-space).
