A question on zero-dimensional spaces I am wondering whether one could see zero-dimensionality of a given compact space just by looking at separable subspaces. To be more precise suppose that we have a compact Hausdorff space $X$ and an uncountable set $S\subset X$. Suppose that for each countable $S_0 \subseteq S$ the closure of $S_0$ is zero-dimensional. Can we conclude that $\overline{S}$ is zero-dimensional as well?
I have worked out the example of $\{0,1\}^\kappa$ ($\kappa>2^\omega$) but I am wondering about the general setting. The space $\{0,1\}^\kappa$ is ccc. What if we assume (to make life easier) that $X$ every closed subspace of $X$ is ccc?
 A: There is at least a consistent counterexample to the conjecture. I’ll sketch the construction. Start with an arbitrary Suslin line $L$; the existence of such a line is consistent with and independent of $\mathsf{ZFC}$. For $x,y\in L$ write $x\sim y$ if the open interval between $x$ and $y$ is countable. Then $\sim$ is an equivalence relation on $L$ whose equivalence classes are order-convex. Thus, the original linear order on $L$ naturally induces a linear order on $L/\sim$, and it’s not hard to see that this order is dense. Moreover, each equivalence class is countable, so $L/\sim$ with its order topology is not separable, and it’s clearly ccc, so it’s a densely ordered Suslin line. Finally, let $X$ be the Dedekind completion of $L/\sim$ together with any missing endpoint; $X$ is a compact, connected Suslin line.
In fact no open interval in $L/\sim$ is separable, so every countable subset of $X$ is nowhere dense and therefore has zero-dimensional closure. Taking $S$ to be $X$ itself gives you a counterexample.
