Prove, if $X$ is compact and $Y$ is paracompact, then $X \times Y$ is paracompact (under product topology) I was wondering if what I've done is valid. I'm new to topology and relatively new to proof-based math other than linear algebra and a bit of one-dimensional calc. Any help would be appreciated.
https://i.stack.imgur.com/Ynr74.png
 A: You’ve gone astray right at the start, I’m afraid: to show that $X\times Y$ is paracompact, you must start with an arbitrary open cover of $X\times Y$, not just one composed of basic open sets of the form $U\times V$ for $U\in\tau_X$ and $V\in\tau_Y$. (And there is no need to tinker with bases for the topologies of $X$, $Y$, and $X\times Y$.)
So let $\mathscr{U}$ be an open cover of $X\times Y$. For each $p=\langle x_p,y_p\rangle\in X\times Y$ there is a $U_p\in\mathscr{U}$ such that $p\in U_p$, and there are open sets $V_p$ in $X$ and $W_p$ in $Y$ such that $p\in V_p\times W_p\subseteq U_p$. Let $\mathscr{B}=\{V_p\times W_p:p\in X\times Y\}$; clearly $\mathscr{B}$ is a basic open refinement of $\mathscr{U}$. Any locally finite open refinement of $\mathscr{B}$ is automatically a locally finite open refinement of $\mathscr{U}$, so now we can work with nice basic open sets in the product.
We first use the compactness of $X$ to get an even nicer refinement of $\mathscr{B}$ by basic open sets in the product. Specifically, for each $y\in Y$ we’ll get a finite subcollection of $\mathscr{B}$ that covers the cross-section $X\times\{y\}$. For each $y\in Y$ let $\mathscr{B}_y=\{V_p:p_y=y\}$; $\mathscr{B}_y$ is an open cover of $X$, and $X$ is compact, so there is a finite $F_y\subseteq X$ such that $\{V_{\langle x,y\rangle}:x\in F_y\}$ covers $X$.
We now have $X\times\{y\}$ covered by the finite family $\{V_{\langle x,y\rangle}\times W_{\langle x,y\rangle}:x\in F_y\}$, where the sets $W_{\langle x,y\rangle}$ are all open nbhds of $y$ in $Y$. There are only finitely many of them, so their intersection is open: if we set $H_y=\bigcap_{x\in F_y}W_{\langle x,y\rangle}$, then $H_y$ is an open nbhd of $y$. Moreover, we can shrink each of the sets $V_{\langle x,y\rangle}\times W_{\langle x,y\rangle}$ for $x\in F_y$ down to $V_{\langle x,y\rangle}\times H_y$ and still cover $X\times\{y\}$; in fact,
$$\bigcup_{x\in F_y}(V_{\langle x,y\rangle}\times H_y)=X\times H_y\;.$$ And these shrunken sets still refine $\mathscr{U}$: $$V_{\langle x,y\rangle}\times H_y\subseteq V_{\langle x,y\rangle}\times W_{\langle x,y\rangle}\subseteq U_{\langle x,y\rangle}$$ for each $x\in F_y$.
We now use the paracompactness of $Y$. The open sets $H_y$ cover $Y$, so $\{H_y:y\in Y\}$ has a locally finite open refinement $\mathscr{G}$. For each $G\in \mathscr{G}$ there is a $y(G)\in Y$ such that $G\subseteq H_{y(G)}$. Let $\mathscr{R}_G=\{V_{\langle x,y(G)\rangle}\times G:x\in F_{y(G)}\}$; then $$\bigcup_{x\in F_{y(G)}}(V_{\langle x,y(G)\rangle}\times G)=X\times G\;,$$ and $$V_{\langle x,y(G)\rangle}\times G\subseteq V_{\langle x,y(G)\rangle}\times H_{y(G)}\subseteq V_{\langle x,y(G)\rangle}\times W_{\langle x,y(G)\rangle}\subseteq U_{\langle x,g(G)\rangle}\;,$$ so $\mathscr{R}=\bigcup_{G\in\mathscr{G}}\mathscr{R}_G$ is an open cover of $X\times Y$ refining $\mathscr{U}$.
All that remains is to show that $\mathscr{R}$ is locally finite, and I’ll leave it to you to finish that off. First verify that $\{X\times G:G\in\mathscr{G}\}$ is a locally finite cover of $X\times Y$, and then use the fact that each $\mathscr{R}_G$ is finite.
The steps that I used to get the sets $H_y$ are useful enough to have been pulled out as a separate result often known as the tube lemma, something that is well worth knowing.
