Proof: Let $\{U_α\}$ be an open cover of $X×Y$. For each $x∈X$, the space $\{x\}×Y≅Y$ is compact, so let it be covered by finitely many $U_{x_1},...,U_{x_n}$ with nontrivial intersection with $\{x\}×Y$, and let $W_x=∩π_1(U_{x_i})$. Then $\{W_x\}_{x∈X}$ is an open cover of $X$, so let $\mathcal{A}$ be a locally finite open refinement covering $X$. For each $A∈\mathcal{A}$, finitely many $U_{x_1},...,U_{x_n}$ cover $A×Y⊆W_x×Y$ (for some $x$), so let $C_A=\{(A×Y)∩U_{x_i}\}$. We claim $\mathcal{C}=∪_{A∈\mathcal{A}}C_A$ is a locally finite open refinement of $\{U_α\}$ covering $X×Y$.
Every element of each $C_A$ is contained in some $U_α$, so $\mathcal{C}$ is clearly a refinement. As well, given $z=x×y∈X×Y$, let $x∈A∈\mathcal{A}$ so that $z∈A×Y=∪C_A$ implying $z$ is contained in an element of $C_A$ and now $\mathcal{C}$ covers $X$. Finally, choose a neighborhood $U$ of $x$ intersecting only finitely many members $A_i∈\mathcal{A}$. Then $U×Y$ is a neighborhood of $z$ that can intersect only among the members of $C_{A_i}$, all of which are finite. Therefore $\mathcal{C}$ is locally finite and $X×Y$ is paracompact.$~\square$
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