Counterexample to show that the class of Moscow spaces is not closed hereditary. What is a counterexample to show that the class of Moscow spaces is not closed hereditary?
(A space $X$ is called Moscow if the closure of every open $U \subseteq X$ is the union of a family of G$‎_{‎\delta‎‎‎}$‎-subsets of $X$.)

This is essentially copied from A. V. Arhangel'skii, Moscow spaces and topological groups, Top. Proc., 25; pp.383-416:

Let $D ( \tau )$ be an uncountable discrete space, and $\alpha D ( \tau )$ the one point compactification of $D ( \tau )$.  Then $D ( \tau )$ is a Moscow space, and $D ( \tau )$ is G$_\delta$-dense in $\alpha D ( \tau )$, while $\alpha D ( \tau )$ is not a Moscow space.
  Indeed, let $U$ be any infinite countable subset of $D ( \tau )$.
  Then $U$ is open in $\alpha D ( \tau )$, and $\overline{U} = U \cup \{ \alpha \}$, where $\alpha$ is the only non-isolated point in $\alpha D ( \tau )$.
  Every G$_\delta$-subset of $\alpha D ( \tau )$ containing the point $\alpha$ is easily seen to be uncountable; therefore, $\overline{U}$ is not the union of any family of G$_\delta$-subsets of $\alpha D ( \tau )$.
  Since $\alpha D ( \tau )$ is a closed subspace of a Tychonoff cube, we conclude that the class of Moscow spaces is not closed hereditary.

