Product of a First Countable Space by a Fréchet Space It's easy to see that the product of two First Countable spaces is First Countable, and it's easy to show that every First Countable space is a Fréchet Space (i.e, if $A \subset X$ and $p \in \bar A$ then there exists a sequence $s_n$ of elements of $A$ such that $s_n \rightarrow p$).
On Alan Dow's "More set-theory for topologists" I saw that there is an nice example of two Fréchet spaces such that their product is not Fréchet. But I was wondering, what if one of them is also First Countable? Will the product be a Fréchet Space?
PS: I don't know if I was supposed to post this question here or in MO. What do you think?
 A: Let $Y=\{p\}\cup(\omega\times\omega)$, where $p\notin\omega\times\omega$, points of $\omega\times\omega$ are isolated in $Y$, and the $$B_n=\{p\}\cup\big([n,\omega)\times\omega\big)$$ for $n\in\omega$ are a local base at $p$. $Y$ is metrizable: it’s homeomorphic to $$\{\langle 0,0\rangle\}\cup\{\langle 2^{-m},2^{-n}\rangle:m,n\in\Bbb N\text{ and }n\ge m\}$$ as a subspace of $\Bbb R^2$.
Let $X=\{q\}\cup(\omega\times\omega)$, where again points of $\omega\times\omega$ are isolated, and the sets $$B_f=\{q\}\cup\{\langle m,n\rangle\in\omega\times\omega:n\ge f(m)\}$$ for $f\in{^\omega\omega}$ are a local base at $q$. It’s not hard to check that $X$ is Fréchet. If $q\in\operatorname{cl}_XA$ for some $A\subseteq\omega\times\omega$, there must be some $n\in\omega$ such that $A\cap\big(\{n\}\times\omega\big)$ is infinite; if
$$A\cap\big(\{n\}\times\omega\big)=\{\langle n,m_k\rangle:k\in\omega\}\;,$$
where $m_k<m_{k+1}$ for each $k\in\omega$, then $\big\langle\langle n,m_k\rangle:k\in\omega\big\rangle$ converges to $q$ in $X$.
Now let 
$$\Delta=\left\{\big\langle\langle m,n\rangle,\langle m,n\rangle\big\rangle:m,n\in\omega\right\}\;.$$
Any open nbhd of $\langle q,p\rangle$ contains a set of the form $B_f\times B_n$ for some $f\in{^\omega\omega}$ and $n\in\omega$, and 
$$\big\langle\langle n,f(n)\rangle,\langle n,f(n)\rangle\big\rangle\in\Delta\cap(B_f\times B_n)\;,$$
so $\langle q,p\rangle\in\operatorname{cl}_{X\times Y}\Delta$. However, $\Delta$ is sequentially closed in $X\times Y$, so $X\times Y$ isn’t even sequential.
