Proof: (a) Suppose X=K1∪K2 where K1,K2⊆X are closed and paracompact. By induction the general case will follow from this one. Suppose {Uα} is an open cover of X. Then {Uα∩K1} and {Uα∩K2} are open covers of K1 and K2 respectively, so let B1={Bβ} and B2={Bγ} be locally finite refinements of the two. We claim C=B1∪B2 is a locally finite refinement of {Uα} covering X. The refinement and covering conditions are evident, so we proceed to demonstrate local finiteness; assume x∉K1. If U is a neighborhood of x such that K2∩U intersects only finitely many members of B2, then U∩(X−K1) is a neighborhood of x intersecting among the same finite family from B2, and is disjoint from the members of B1. So we may assume x∈K1, and similarly x∈K2. As such let U and V be neighborhoods of x such that K1∩U and K2∩V intersect only finitely many members of B1 and B2 respectively. If U∩V intersects an element of B1, then since that element is within K1 so too does U∩V∩K1⊆U∩K1, so that only finitely many elements of B1 are intersected. Similarly too for B2, and now U∩V is a neighborhood of X intersecting only finitely many elements of B.
(b) Let X=∪int Ki where each Ki is paracompact. Again let {Uα} be an open cover of X. For each i, let Bi be a locally finite refinement of {Uα∩Ki} covering Ki, and further let Ai={B∩int Ki | B∈Bi}. Then Ai is a locally finite open cover of int Ki for each i, so that A=∪Ai is a countably locally finite open refinement of {Uα} covering X, so that X is paracompact by Lemma 41.3. ◻
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