Choquet not strong Choquet

Question

Is there a uncountable Hausdorff Choquet space which is not strong Choquet?

I know a counterexample of a Choquet non strong Choquet space, but is countable non Hausdorff.

Let $X$ be the disjoint union of $\omega$ and $\mathbb{Q}$. Equip $X$ with the topology generated by the base:

$\{\{n\}:n∈ω\}∪\{(a,b)∪C:a,b∈\mathbb{Q}\}$, with C a cofinite subset ofω}. $X$ is Choquet non strong Choquet space.

Answer 1

There even is a metric space $X$ with the required properties.

Let $X=(\Bbb R \times [0,\infty)) \setminus \{(x,0): x \text{ irrational }\}$ in the subspace topology of the plane, is an uncountable metric (so Hausdorff) space.

In Kechris' book (Descriptive Set Theory) it is shown that a co-meagre subset of a Polish space (as $X$ is) is Choquet. But it is not strongly Choquet as it's not completely metrisable (as it contains a closed embedded copy of $\Bbb Q$).