A space is Lindelöf if every open cover admits a countable subcover.
A space is Menger if given a countable sequence of open covers $\mathcal U_n$ for $n<\omega$ there exist finite subcollections $\mathcal F_n\subseteq\mathcal U_n$ such that $\bigcup\{\mathcal F_n:n<\omega\}$ is a cover.
Is every Lindelöf space Menger?
(The answer to this question is asserted in the comments of this MO question – I'm posting here as I don't believe it's been answered on the StackExchange network directly.)
Best Answer
No. Consider the Baire space $\omega^\omega$, the product of countably-many copies of $\omega$ with the discrete topology. This space is second-countable and therefore Lindelöf.
Let $\mathcal U_n=\{[t]:t\in\omega^{\leq n}\}$, where $[t]=\{f\in\omega^\omega:f\text{ extends }t\}$. Then given any choices $\mathcal F_n\subseteq\mathcal U_n$ for $n<\omega$, we will find $f\in\omega^\omega$ that's not covered. First pick $f(0)<\omega$ such that $[\langle f(0)\rangle]\not\in\mathcal F_0$. Then if $f(0),\dots,f(n)$ have been chosen such that $[\langle f(0),\dots,f(i)\rangle]\not\in\mathcal F_i$ for each $i\leq n$, we may also choose $f(n+1)<\omega$ such that $[\langle f(0),\dots,f(n),f(n+1)\rangle]\not\in\mathcal F_{n+1}$.
Now consider this $f$. If $f$ extended $t$ for some $[t]\in\mathcal F_n$, this would violate the choice made for $f(n)$. Thus $f\not\in [t]$ for every $[t]\in\mathcal F_n$, so $f$ is not covered. Thus the space is not Menger.