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@AlexanderOsipov: There can be. In the modified argument that I just posted, which gets rid of the requirement that $\mathrm{non}(\mathcal M)$ be larger than $\mathfrak{b}$, the set $\mathbb Q \setminus C$ is relatively $G_\delta$ in $X$, and it is dense in $X$ as well.
Actually, thinking about this a bit more, I think it's possible to show that $\mathfrak{b} = \aleph_1$ if there is an uncountable separable metric space concentrating on a countable set. So consistency is the best you can hope for. (My guess is that $\mathfrak{b} = \aleph_1$ is the optimal hypothesis -- the other thing is just to make the "meager" part easy.) I will come back tomorrow to write out the details of this, but I can't at the moment because I'm about to attend a seminar, and then head home for the day :)
Nice! Regarding your last sentence, the answer to Dominic's question is yes if $\mathfrak{a} = \mathfrak{c}$. This is because if $\mathcal M$ is a MAD family on $\mathbb N$ and $A \subseteq \mathbb N$, then either $A$ only meets finitely many members of $\mathcal M$ in an infinite set, or else the restriction of $\mathcal M$ to $A$ is a MAD family on $A$, hence of cardinality $\mathfrak{c}$. So it seems the answer to Dominic's question is yes if and only if $\mathfrak{a} = \mathfrak{c}$.
I'm confused by your claim that there can exist a "greatest" $\Pi^1_1$ set without the perfect set property. Given any $\Pi^1_1$ set without the perfect set property, you can add countably many points to it, and it will still be a $\Pi^1_1$ set without the perfect set property. Right?
@Lorenzo: No, I mean the union of all rational shifts of $C$. This won't be a disjoint union; it will give you a dense, $\sigma$-compact, meager subset of $\mathbb R$.
For (3), one example is $\mathbb Q + C$, where $C \subseteq \mathbb R$ is the Cantor set. This is nowhere completely metrizable (no open subset is a relative $G_\delta$, by Baire's theorem), but is $\sigma$-compact. So it is an $F_\sigma$ subspace of any completely metrizable space containing it, which means its complement is completely metrizable. (And the same argument gives a positive answer for question (2).)
@TimCampion: I suppose I was taking it as intuitively obvious. For evidence in my favor, I just asked my 4-year-old daughter, and she said that she knows what it means. I would count my previous comment as a conceptually clear explanation (to me, anyway), but I would not necessarily find it easy to turn the explanation into a rigorous proof.
