TITLE: Topological counterexample: compact, Hausdorff, separable space which is not first-countable
QUESTION [7 upvotes]: I need an example for a compact, Hausdorff, separable space which is not first-countable.
I tried to look for it for some time without success...

REPLY [11 votes]: The space $I^I$ (i.e., product of $\mathfrak c$-many copies of the unit interval $I=[0,1]$ is a compact Hausdorff space.
It is not first-countable, see here: Uncountable Cartesian product of closed interval
It is separable by Hewitt-Marczewski-Pondiczery theorem,
see here: On the product of $\mathfrak c$-many separable spaces
As pointed out in a comment, we could also prove separability by directly showing that polynomials with rational coefficients form a countable dense subset. See also this answer for a similar approach in a slightly different space.