Product of sequential sequentially compact spaces is sequential

Question

I am trying to show that the product of two sequentially compact sequential spaces is sequential. Can someone help me?

Edit: I found that there is a reference for this:

Boehme T.K., Linear s-spaces, Proc. Symp. Convergent structures, Univ. Oklahoma, 1965

However, my university doesn't own a copy, I couldn't find it in the internet and this text isn't on mathscinet!

Edit2: Instead of supposing that both $X$ and $Y$ are both sequentially compact, what if we suppose that both spaces are locally sequentially compact? (in the sense that for every $x \in X$ and every open set $U$ such that $x \in U$ there exists a sequentially compact neighborhood $V$ of $x$ such that $V \subset U$).

Answer 1

Completely revised.

Lemma. Let $X$ be sequential and $Y$ sequentially compact, and let $\pi_X:X\times Y\to X$ be the projection map. If $A\subseteq X\times Y$ is sequentially closed, then $\pi_X[A]$ is closed in $X$.

Proof. Let $P=\pi_X[A]$, and suppose that $P$ is not closed. $X$ is sequential, so $P$ is not sequentially closed. Thus, there are a point $x\in X\setminus P$ and a sequence $\langle x_n:n\in\Bbb N\rangle$ in $P$ that converges to $x$. For each $n\in\Bbb N$ there is a $y_n\in Y$ such that $\langle x_n,y_n\rangle\in A$. $Y$ is sequentially compact, so (by passing to a subsequence if necessary) we may assume that $\langle y_n:n\in\Bbb N\rangle$ converges to some $y\in Y$. But then $\big\langle\langle x_n,y_n\rangle:n\in\Bbb N\big\rangle$ is a sequence in $A$ converging to $\langle x,y\rangle$, so $\langle x,y\rangle\in A$, and $x\in P$, contradicting the choice of $x$. $\dashv$

Now suppose that $X$ is sequential, $Y$ is sequential, regular, and locally sequentially compact, and that $A\subseteq X\times Y$ is sequentially closed but not closed. Fix $p=\langle x,y\rangle\in(\operatorname{cl}A)\setminus A$, and let $A_x=\big(\{x\}\times Y\big)\cap A$. $A_x$ is sequentially closed, and $Y$ is sequential, so $A_x$ is closed. Thus, $p$ has an open nbhd $U$ disjoint from $A_x$. Let $C$ be a sequentially compact nbhd of $y$ such that $\{x\}\times C\subseteq U$, and let $A_0=A\cap(X\times C)$; $A_0$ is sequentially closed, $p\in(\operatorname{cl}A_0)\setminus A_0$, and $A_0\cap A_x=\varnothing$.

Let $P=\pi_X[A_0]$, where $\pi_X:X\times Y\to X$ is the projection map; it follows from the lemma (applied to $X\times C$) that $P$ is closed in $X$. But then $(X\setminus P)\times Y$ is an open nbhd of $p$ disjoint from $A_0$, which is impossible. Thus, $A$ must in fact be closed, and $X\times Y$ must be sequential.