I'm trying to prove any manifold is paracompact, and I'm stuck in the statement marked in red ink. Could anyone tell me how to arrive at this observation? Thank you.
If we add the hypothesis of second-countability to a locally compact Hausdorff space, we can prove even more. A sequence $(K_i)_{i=1}^\infty$ of compact subsets of a topological space $X$ is called an exhaustion of $\boldsymbol{X}$ by compact sets if $X=\bigcup_iU_i$ and $K_i\subseteq\mathrm{Int}\,K_{i+1}$ for each $i$.
Proposition A.60. A second-countable, locally compact Hausdorff space admits an exhaustion by compact sets.
Theorem 1.15 (Manifolds Are Paracompact). Every topological manifold is paracompact. In fact, given a topological manifold $M$, an open cover $\mathscr{X}$ of $M$, and any basis $\mathscr{B}$ for the topology of $M$, there exists a countable, locally finite open refinement of $\mathscr{X}$ consisting of elements of $\mathscr{B}$.
Proof. Given $M$, $\mathscr{X}$, and $\mathscr{B}$ as in the hypothesis of the theorem, let $(K_j)_{j=1}^\infty$ be an exhaustion of $M$ by compact sets (Proposition A.60). For each $j$, let $V_j=K_{j+1}\setminus\mathrm{Int}\,K_j$ and $W_j=\mathrm{Int}\,K_{j+2}\setminus K_{j-1}$ (where we interpret $K_j$ as $\varnothing$ if $j<1$). Then $V_j$ is a compact set contained in the open subset $W_j$. For each $x\in V_j$, there is some $X_x\in\mathscr{X}$ containing $x$, and because $\mathscr{B}$ is a basis, there exists $B_x\in\mathscr{B}$ such that $x\in B_x\subseteq X_x\cap W_j$. The collection of all such sets $B_x$, as $x$ ranges over $V_j$ is an open cover of $V_j$, and thus has a finite subcover. The union of all such finite subcovers as $j$ ranges over the positive integers is a countable open $\color{red}{\text{cover of }M}$ that refines $\mathscr{X}$. Because the finite subcover of $V_j$ consists of sets contained in $W_j$, and $W_j\cap W_{j'}=\varnothing$ except when $j-2\leq j'\leq j+2$, the resulting cover is locally finite. $\qquad\square$
