Is this Michael subspace $M$ submetrizable? For a more detailed discussion of Bing’s Example G in this blog, see the blog post Bing's example G. For the sake of completeness, we repeat the definition of Example G. Let $Q$ be the set of all subsets of $P$. Let $F=2^Q$ be the set of all functions $f: Q\to 2=\{0,1\}$. Obviously $2^Q$ is simply the Cartesian product of $|Q|$ many copies of the two-point discrete space $\{0,1\}$, i.e., $\prod_{q\in Q}\{0,1\}$. For each $p \in P$, define the function $f_p: Q\to 2$ by the following
$$
\forall q \in Q, f_q(p)=1, \text{ if } p \in q \text{ and } f_q(p)=0, \text{ if } p \notin q
$$
Let $F_P=\{f_p: p\in P\}$. Let $\tau$ be the set of all open subsets of $2^Q$ in the product topology. The following is another topology on $2^Q$:
$$
\tau^*=\{ U\cup V: U \in \tau \text{ and } V \subset 2^Q \text{ with } V \cap F_P=\emptyset \}
$$
Bing’s Example G is the set $F=2^Q$ with the topology $\tau^*$. In other words, each $x \in F\setminus F_P$ is made an isolated point and points in $F_P$ retain the usual product open sets.

Michael’s Subspace of Example G 

For each $f\in F$, let $supp(f)$ be the support of $f$, i.e., $supp(f)=\{q\in Q: f(q)\not=0 $. Michael considered the following subspace of $F$:

$$ M=F_p\cup \{f \in F: supp(f) \text{ is finite }\}.$$
My question is this:
Is this Michael subspace $M$ submetrizable?
Thanks for your help.
 A: The claim made in the link that $M$ is perfectly normal is false, I think. I have found a mistake in the proof (I contacted the author via a comment) and I think the following shows that $M$ cannot be perfectly normal (references are to Burke's chapter in the Handbook of Set-theoretic topology):
Suppose $M$ is perfectly normal, then so are all subspaces. By theorem 4.9(ii) the subspace $Y$ of all points $f$ that have at most two singletons sets (members of $Q$!) at which they have the value 1 is normal and metacompact (so weakly $\theta$-refinable) but not subparacompact. By the assumption, $Y$ is also perfectly normal. But theorem 4.17 (Lutzer and Bennett) says that a perfect weakly $\theta$-refinable is subparacompact, contradiction. A simpler, more direct argument is probably possible.
Points are $G_\delta$ sets, I think: the set $F_n$ of all points in $F$ with support of size at most $n$ is closed in $F$, so $(F \setminus F_n) \cap M$ is open in $M$ and it has intersection $F_P$. For some $p \in P$, also intersecting with an open set $W_p$ that witnesses the discreteness of $F_P$ for $f_p$, shows that each $\{f_p\}$ is a $G_\delta$ (but there is no countable local base at such points, I don't think). So this does not exclude yet that $M$ is submetrizable. I have a hunch it probably is not, but no proof yet.    
