Let $\ C\ $ be an uncountable compact subspace of $\ \mathbf R.\ $ Let W be the family of all open (in $\ C\ $) countable sets, and $\ V := \bigcup W.\ $ Family $\ W\ $ admits a countable covering of $\ V.\ $ Thus, the union $\ V\ $ is countable. Now, let's replace $\ C\ $ by $\ C\setminus V.\ $ The new $\ C\ $ is still uncountable. Furthermore, every nonempty open subset of $\ C\ $ is uncountable.

Let $\ K\ $ be the linearly ordered set obtained from $\ C\ $ by identifying every two $\ a\ b \in C\ $ such that $\ (a;b)\cap C=\emptyset.\ $ Set $\ K\ $ is still uncountable (there are only countably many points $\ x\in\ C\ $ which belong to classes which have more than one point). The topological space canonically associated with $\ K\ $ has the minimal and maximal point, and it is easy to check that $\ K\ $ has Dedekind property (for an interval) induced by Dedekind property of $\ \mathbf R.\ $ Thus $\ K\ $ is a connected Hausdorff compact space.

(There is a Urysohn function of $\ K\ $ onto a non-trivial interval).