This week's question was correctly answered by caffeinemachine and Opalg.
Here's Opalg's solution: [sp]Suppose that $X$ is Lindelöf, $Y$ is compact, and let $\mathcal{U} = \{U_\alpha\}$ be an open cover of $X\times Y$. Fix $z\in X$. For each $y\in Y$ the point $(z,y)$ belongs to some $U_\alpha$. Let $L_y\times M_y$ be a basic open neighbourhood of $(z,y)$ contained in $U_\alpha$ (so that $L_y$ is an open neighbourhood of $z$ in $X$, $M_y$ is an open neighbourhood of $y$ in $Y$, and $L_y\times M_y \subseteq U_\alpha$). Then $\{M_y\}$ is an open cover of $Y$ and so has a finite subcover $\{M_1, \ldots,M_n\}.$ Let $\{L_1, \ldots,L_n\}$ be the corresponding open subsets of $X$ and let $\{U_1,\ldots, U_n\}$ be the corresponding sets in $\mathcal{U}$, so that $L_j\times M_j\subseteq U_j$. Let $\mathcal{U}_z = \{U_1,\ldots,U_n\}.$
Let $$V_z = \bigcap_{1\leqslant j\leqslant n} L_j$$. Then $V_z$ is an open neighbourhood of $z$, and $$(x,y) \in \bigcup_{U\in \mathcal{U}_z}U$$ for all $(x,y)\in V_z\times Y.$
Now let $z$ vary. The sets $V_z$ form an open cover of $X$ and therefore have a countable subcover $\{V_{z_k}\}$ say. For each $k$, the set $\mathcal{U}_{z_k}$ is finite, so that $\bigcup_k \mathcal{U}_{z_k}$ is countable. Finally, every element of $X\times Y$ is in one of those sets, so they form a countable subcover of $\mathcal{U}.$ This shows that $X\times Y$ is Lindelöf.[/sp]
Here's another solution, which makes use of the
tube lemma.
[sp]Let $\mathcal{U}$ be an open cover of $X\times Y$. Then for each $x\in X$, $\mathcal{U}$ is also an open cover for $\{x\}\times Y$. Since $Y$ is compact, there exists for each $x\in X$ a finite subcollection $\mathcal{U}_x$ of $\mathcal{U}$ which covers $\{x\}\times Y$. Let $U_x$ be the set obtained by unioning the elements of $\mathcal{U}_x$. This set is open in $X\times Y$ and contains $\{x\}\times Y$. Once again, using the fact that $Y$ is compact, it follows by the tube lemma that for each $x\in X$, there exists an open neighborhood $N_x$ of $x$ such that $N_x\times Y \subset U_x$. Now, consider the collection $\{N_x:x\in X\}$. Since $X$ is Lindelöf, there exists a countable subset $I\subset X$ such that $\{N_x:x\in I\}$ covers $X$. Thus, we see that $\bigcup_{x\in I} \mathcal{U}_x$ is a countable subcollection of $\mathcal{U}$ that covers $X\times Y$. Therefore, $X\times Y$ is Lindelöf.[/sp]