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

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).

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

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

Let $\ C\ $ be an uncountable compact subset of $\ \mathbf R.\ $ Let's assume that $\ C\ $ has no isolated points (if necessary, replace $\ C\ $ by $\ C\ $ minus all its isolated points).

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).

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