A Question about Productively Lindelöf Metrizable Spaces I've been reading more productively Lindelöf spaces and the Michael Space Problem. The Michael Space Problem asks if there exists a Michael Space.

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*A topological space $X$ is productively Lindelöf if it's product with any Lindelöf space $Y$ is also Lindelöf.

*A topological space $X$ is a Michael space if it's Lindelöf, but it's product with $\mathbb{P}$ (the set of irrational numbers) is not Lindelöf.

In an article by F.D. Tall, he states:

The Continuum Hypothesis (CH) implies that productively Lindelöf metrizable spaces are $\sigma$-compact.

He also mentions that this is implicitly proved in E. Michael's article, and explicitly stated and proved in an article by K. Alster, but I am not able to access Alster's article.
Michael's article shows how CH can be used to create a Michael space, that is:

CH implies there exists a Lindelöf space $X$ such that $X \times \mathbb{P}$ is not Lindelof.

I've read through the proofs in Michael's article, but I'm not sure how it can be implicitly shown that Lindelöf metrizable spaces are $\sigma$-compact if CH is assumed.
Any insight would be greatly appreciated.
 A: For the record, Alster’s paper is available from this page. The relevant result (slightly restated) is

Proposition $\mathbf{1}$**:** Assume that the Continuum Hypothesis holds. Then if $Z\subseteq\Bbb I^A$ and $Z$ is productively Lindelöf, then for every countable $S\subseteq A$ the space $p_S[Z]$ is $\sigma$-compact, where $p_S$ denotes the projection from $\Bbb I^A$ onto $\Bbb I^S$. 

If $Z$ is metrizable, then $Z$ embeds in $\Bbb I^\omega$, $Z$ itself is $\sigma$-compact.
The proof isn’t too bad. Suppose not, and let $S\subseteq A$ be countable and such that $p_S[Z]$ is not $\sigma$-compact. Let $B=\Bbb I^S\setminus p_S[Z]$; if $B$ were a $G_\delta$ in $\Bbb I^S$, $p_S[Z]$ would be $\sigma$-compact, so $B$ is not a $G_\delta$ in $\Bbb I^S$. By $\mathsf{CH}$ there are $\omega_1$ open sets in $\Bbb I^S$ containing $B$, say $\{U_\xi:\xi<\omega_1\}$. For $\eta<\omega_1$ let $G_\eta=\bigcap_{\xi\le\eta}U_\xi$; then $\langle G_\xi:\xi<\omega_1\rangle$ is a non-increasing $\omega_1$-sequence of $G_\delta$ sets in $\Bbb I^S$ such that if $U$ is any open nbhd of $B$ in $\Bbb I^S$, there is a $\xi<\omega_1$ such that $B\subseteq G_\xi\subseteq U$. Since $B$ is not itself a $G_\delta$ in $\Bbb I^S$, $G_\xi\cap p_S[Z]$ is uncountable for each $\xi<\omega_1$, so for each $\xi<\omega_1$ we can choose a point $x_\xi\in G_\xi\cap p_S[Z]$ in such a way that the points $x_\xi$ are distinct.
Let $X=B\cup\{x_\xi:\xi<\omega_1\}$, and refine the subspace topology that it inherits from $\Bbb I^S$ by making each $x_\xi$ an isolated point; let $\tau$ be the resulting topology. Let $\mathscr{V}\subseteq\tau$ be an open cover of $X$. $\Bbb I^S$ is compact and metrizable, so it’s hereditarily Lindelöf, and there is a countable $\mathscr{V}_0\subseteq\mathscr{V}$ that covers $B$. By construction $X\setminus\bigcup\mathscr{V}_0$ is countable and hence Lindelöf, so $X$ is Lindelöf. However, $\{\langle x_\xi,x_\xi\rangle:\xi<\omega_1\}$ is a closed discrete subset of $X\times p_S[Z]$, which is therefore not Lindelöf, contradicting the fact that productive Lindelöfness is preserved by continuous maps. $\dashv$
I’ve not gone through the details, but this is basically derived from the construction of Michael’s Example $1.2$.
