QUESTION: Does every regular perfect space with the countable chain condition have cardinality bounded above by the continuum? Is this at least true for perfectly normal ccc spaces?
Recall that a space is perfect if every closed set is a $G_\delta$ set (that is, a countable intersection of open sets). We say that a space has the countable chain condition if every family of pairwise disjoint non-empty open sets is countable. Perfectly normal just means perfect and normal.
If one strengthens the ccc to the condition every discrete set is countable then the answer is yes. This follows from Hajnal and Juhasz's result that every space where singletons are $G_\delta$ and discrete sets are countable has cardinality at most continuum. Regularity is not needed for this to be true ($T_1$ is enough).
(It's easy to prove that in a perfect space where closed discrete sets are countable, it's even true that every discrete set is countable. So from the above statement it follows that every Lindelof $T_1$ perfect space has cardinality at most the continuum, something that was proved by Alexandroff and Urysohn in their Memoire, at least for compact Hausdorff spaces, if I'm not mistaken).
On the other hand:
There are Hausdorff ccc perfect spaces of arbitrarily large cardinality.
For example, let $\kappa$ be any cardinal and $D_n=\{x \in 2^\kappa: |x^{-1}(1)|=n \}$. Set $X=\bigcup_{n<\omega} D_n$. Let $\tau$ be the refinement of the usual topology on $X$ obtained by making every $D_n$ closed discrete. In other words, a basic open set has the form $U \setminus \bigcup_{n\in F} D_n$ where $U$ is open in the topology on $X$ inherited from $2^\kappa$ and $F \subset \omega$ is finite. This space is ccc: indeed, let $\{U_\alpha: \alpha<\aleph_1\}$ be a pairwise disjoint family of open sets of cardinality $\aleph_1$. By the pigeonhole principle we can assume that for some fixed finite set $F$ we have $U_\alpha=V_\alpha \setminus \bigcup_{n \in F} D_n$, where $V_\alpha$ is open in the usual topology on $X$, for every $\alpha<\aleph_1$. So for all $\alpha, \beta <\aleph_1$ such that $\alpha \neq \beta$ we have that $V_\alpha \cap V_\beta \subset \bigcup_{n \in F} D_n$ which implies that $V_\alpha \cap V_\beta$ is empty, as $\bigcup_{n\in F} D_n$ is nowhere dense. But this is a contradiction since $X$ with the topology inherited from $2^\kappa$ is dense in the ccc space $2^\kappa$ and thus it is also ccc.
The space $(X, \tau)$ is more than perfect. As a matter of fact, let $G \subset X$ be any set and set $G_n=D_n \setminus G \cap D_n$. Note that $G_n$ is closed. Then $G=\bigcap_{n<\omega} (X \setminus G_n)$, which proves that $G$ is a $G_\delta$ set.
