A compact set in a noncompact manifold contained in a union of two compact sets

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Recently I'm reading Hatcher's text (Edition 2002) and stuck in the discussion of Poincare duality for noncompact manifolds (p.245-248). More detailly, there is an argument in the proof of lemma 3.36 that I can not yet work out. A screenshot of Lemma 3.36

From the picture, you can find a sentence below the second diagram:

Since each compact set in $U\cap V$ is contained in an intersection $K\cap L$ of compact sets $K\subset U$ and $L\subset V$ , and similarly for $U\cup V$ , the diagram induces a limit diagram having the form stated in the lemma.

The case of intersection is trivial, just taking $K$ and $L$ both to be the given compact set. But it's hard for me to deal with the union's case. My first attempt is to take $K$ to be $U\cap B$ and $L$ to be $V\cap B$, where $B$ is the compact set in $U\cap V$. Then I tried to verify if they were compact in $U$ and $V$ respectively, but I failed. I have been confused for a long time. Is there anyone who knows how to construct these two compact sets?

Reminder: In this lemma, $M$ is assumed to be an $R$-orientable $n$-manifold, possibly noncompact. So basically it's Hausdorff (, but there're no more restrictions in the context such as axioms of countability). And $U$, $V$ are two open subsets such that $M=U\cup V$.

Thanks in advance.

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Manifolds are locally compact Hausdorff spaces. So let more generally $X$ be locally compact Hausdorff, $U,V \subset X$ be open and $C \subset U \cup V$ be compact. Each $x \in C \cap U$ has a open neighborhood $U_x$ such that $\overline U_x$ is compact and $\overline U_x \subset U$. Similarly each $y \in C \cap V$ has a open neighborhood $V_y$ such that $\overline V_y$ is compact and $\overline V_y \subset V$. Since $C$ is covered by the $U_x$ with $x \in C \cap U$ and the $V_y$ with $y \in C \cap V$, we find finitely many $x_1,\ldots,x_m \in C \cap U$ and finitely many $y_1,\ldots,y_m \in C \cap V$ such that $C \subset \bigcup_{i=1}^m U_{x_i} \cup \bigcup_{j=1}^n V_{y_j}$. Now $K = \bigcup_{i=1}^m \overline U_{x_i}$ and $L = \bigcup_{j=1}^n \overline V_{y_j}$ are compact and we have $K \subset U, V \subset V$ and $C \subset K \cup L$.