Is a separable compact Hausdorff space perfectly normal? 
Let $X$ be a separable compact Hausdorff topological space. Then every closed subset of $X$ is a $G_{\delta}$ set (i.e. $X$ is perfectly normal).

I've been skimming through some topology textbooks recently. As we know, every closed subset of a metric space is a $G_{\delta}$ set. But I am not sure whether the above proposition is correct because I did not find any counterexample. Can anyone give a proof or counterexample? Thanks.
 A: Let $X = \{0,1\}^{\mathbb{R}}$. Since $\{0,1\}$ is separable and $\operatorname{card} (\mathbb{R}) \leqslant 2^{\aleph_0}$, $X$ is separable. Let
$$F = \{0\} = \{ x \in X : t \in \mathbb{R} \implies x(t) = 0\}.$$
$F$ is a closed subset of $X$ that is not a $G_{\delta}$-set. For if $U$ is an open subset of $X$ containing $F$, then there is a finite subset $I = I(U)$ of $\mathbb{R}$ such that
$$V_I = \{ x \in X : t \in I \implies x(t) = 0\} \subset U.$$
Then if we have a countable family $\{ U_n : n \in \mathbb{N}\}$ of open subsets containing $F$, the union
$$K := \bigcup_{n\in \mathbb{N}} I(U_n)$$
is a countable, thus proper, subset of $\mathbb{R}$, and hence
$$\bigcap_{n\in \mathbb{N}} U_n \supset \bigcap_{n \in \mathbb{N}} V_{I(U_n)} = V_K \supsetneqq F.$$
A: Although Henno’s general result renders specific examples a bit superfluous, I’ll note that $\beta\Bbb\omega$ is another relatively easy example. It is certainly compact, Hausdorff, and separable. The ultrafilter construction of $\beta\Bbb\omega$ makes it obvious that it has a base of cardinality $2^\omega=\mathfrak{c}$: if for each $A\subseteq\omega$ we set
$$\widehat A=\{p\in\beta\omega:A\in p\}\;,$$
then $\mathscr{B}=\left\{\widehat A:A\subseteq\omega\right\}$ is a base for $\beta\omega$. 
If $p\in\beta\omega$, and $\{p\}$ is a $G_\delta$, there is a countable $\mathscr{B}_p\subseteq\mathscr{B}$ such that $\bigcap\mathscr{B}_p=\{p\}$. $\mathscr{B}$ has only $\left(2^\omega\right)^\omega=2^\omega$ countable subsets, so there can be at most $2^\omega$ points $p\in\beta\omega$ such that $\{p\}$ is a $G_\delta$. But $|\beta\omega|=2^{\mathfrak{c}}$, so most singletons in $\beta\omega$ are closed sets that are not $G_\delta$ sets.
