The question was very popular over on MSE but seems to have left everyone speechless. Maybe someone here can help?

Definition:Suppose $X$ is a compact connected Hausdorff space and $D \subset X$ countable and dense. We say $D$ isdivisibleto mean for every open $U \subset X$ there exists a partition $D \cap U = D_1 \cup D_2$ with $D_1$ and $D_2$ both dense in $U$. We call $X$divisibleto mean it is separable and each countable dense subset is divisible.

Is every compact connected Hausdorff space divisible?

The property is easy to show if $X$ is metric rather than just Hausdorff. Just let $U_1,U_2, \ldots $ be a countable basis. Since $X$ is connected each $U_n$ contains a proper subcontinuum which surjects onto $[0,1]$ by Urysohn's lemma. Hence each open subset is uncountable. So choose distinct $a_1,b_1 \in U_1$ and proceed by induction. At each stage we have selected only finitely many $a_n,b_n$ and $U_m$ contains infinitely many other elements. So select distinct $a_m,b_m \in U_m - \{a_1,b_2, \ldots , a_{m-1},b_{m-1}\}$ and define $D_1 = \{a_1,a_2,\ldots\}$ and $D_2 = \{b_1,b_2,\ldots\} \cup (D - D_1)$. Then $D_1 \cap U$ and $D_2 \cap U$ form the desired partition for any choice of $U$. So $X$ might be called *simultaneously divisible*.

The property fails for (disconnected) spaces with an isolated point. Because if $\{x\} \subset X$ is open $x$ must be an element of exactly one of $D_1,D_2$ which means the other is not dense. I am not sure if non-divisible spaces are any easier to come by if we only perfect spaces instead of connected spaces.

Under AC being Hausdorff is necessary. For the cofinite topology is compact and connected and any infinite subset is dense. Without AC $D$ might be an amorphous set.

resolvability. There is a closely related question here (different, but related): mathoverflow.net/questions/275424/…. $\endgroup$ – Will Brian Apr 17 '18 at 13:03separable. Strange. Can you say more? $\endgroup$ – Wlod AA Apr 18 '18 at 5:20