There are old ZFC examples due to Eric van Douwen that satisfy all the properties in the title, except that they are of cardinality $2^{\aleph_0}$, so the answer to the title question is YES if the Continuum Hypothesis (CH) is assumed, but I am asking for an example which does not use any axiom beyond the ZFC axioms.

It would be especially nice if it were first countable, like Eric's examples, and normal, like one of his examples. A Souslin tree with the interval topology qualifies, but the existence of Souslin trees is ZFC-independent. I have been able to weaken CH to "stick" [which says that there is a family of $\aleph_1$ countable subsets of an uncountable set, such that every uncountable subset contains a member of the family] and I also have an example if $\mathfrak b = \aleph_1$ but no ZFC example.

It would also be interesting to know whether the existence of an example implies the existence of a first countable example. There are lots of first countable examples under CH, and the "stick" example I have in mind is also first countable, as is my $\mathfrak b = \aleph_1$ example. For sure, there is a scattered example if there is one at all: the "Kunen line" qualifies under CH, while if CH is negated, then we use the fact that every crowded (= dense-in-itself) compact Hausdorff space is of cardinality at least $2^{\aleph_0}$.