Is a hausdorff perfect space which is not first countable neccessarily uncountable?

Question

Counterexamples in Topology has a couple of countable spaces which aren't first countable, but none of them are perfect spaces. I'm looking either for a theorem that says that such a space can't exist or a counter-example.

Answer 1

Such a space can be countable.

Let $p$ be a free ultrafilter on $\Bbb N$, let $Y=\Bbb Q\times\Bbb N$, and let $X=\{p\}\cup Y$. Make $Y$ an open subset of $X$ with its product topology. Basic open nbhds of $p$ are sets of the form $\{p\}\cup(\Bbb Q\times U)$ for $U\in p$. The resulting space is countable and Hausdorff, has no isolated points, and is not first countable at $p$.

Answer 2

Let $X$ be the product of continuum many copies of $[0,1]$. Then $X$ is a compact Hausdorff space; $X$ is separable, because it's the product of continuum many separable spaces; and $X$ is not first countable at any of its points.

Let $Y$ be a countable dense subspace of $X$. Then $Y$ is a Tychonoff space, so in particular a Hausdorff space; $Y$ has no isolated points because it's a dense subspace of $X$, which has no isolated points; and $Y$ is not first countable at any of its points, because $X$ is a regular space which is not first countable at any of its points, and $Y$ is a dense subspace of $X$.