A confusion about Lemma 16.2 in Munkres' Analysis on Manifolds

In Munkres's Analysis on Manifolds, page 137 Lemma 16.2, it states: Step 1. Let $D_{1},D_{2},\cdots$ be a sequence of compact (rectifiable) subsets of (the open set) $A$ whose union is $A$, such that $D_{i}\subset\mathrm{Int}D_{i+1}$ for each $i$. For convenience in notation, let $D_{i}$ denote the empty set for $i\leq0$. Then for each $i$, define $B_{i}=D_{i}-\mathrm{Int}D_{i-1}$. The set $B_{i}$ is bounded, being a subset of $D_{i}$; and it is closed, being the intersection of closed sets $D_{i}$ and $\mathbf{R}^{n}-\mathrm{Int}D_{i-1}$. Thus $B_{i}$ is compact. Also, $B_{i}$ is disjoint from the closed set $D_{i-2}$, since $D_{i-2}\subset\mathrm{Int}D_{i-1}$.

Munkres just uses the fact that $B_{i}$ is disjoint from the closed set $D_{i-2}=\bigcup_{k=1}^{i-2}B_{k}$ to continue his proof. Then for all $\mathbf{x}\in B_{i}$, there exists $\epsilon_{\mathbf{x}}>0$, such that $C(\mathbf{x},\epsilon_{\mathbf{x}})$ is disjoint from the closed set $D_{i-2}=\bigcup_{k=1}^{i-2}B_{k}$...

However, I think we can also get $B_{i}$ is disjoint from the set $D_{i+2}^{\ast}=\bigcup_{k=i+2}^{\infty}B_{k}$. It seems that for all $\mathbf{x}\in B_{i}$, there exists $\epsilon_{\mathbf{x}}>0$, such that $C(\mathbf{x},\epsilon_{\mathbf{x}})$ is disjoint from the set $D_{i+2}^{\ast}=\bigcup_{k=i+2}^{\infty}B_{k}$. The difficulty arises from the fact that the set $D_{i+2}^{\ast}=\bigcup_{k=i+2}^{\infty}B_{k}$ is a union of countable many closed set $B_{k}$, which may not be closed. So, how can we prove/disprove for all $\mathbf{x}\in B_{i}$, there exists $\epsilon_{\mathbf{x}}>0$, such that $C(\mathbf{x},\epsilon_{\mathbf{x}})$ is disjoint from the set $D_{i+2}^{\ast}=\bigcup_{k=i+2}^{\infty}B_{k}$?

Here is my try. $B_{i}$ is disjoint from the closed set $B_{i+2}$. Choose any $\mathbf{x}\in B_{i}$. There exists $\epsilon_{\mathbf{x}}>0$, such that $C(\mathbf{x},\epsilon_{\mathbf{x}})$ is disjoint from the closed set $B_{i+2}$. We just need to prove if $C(\mathbf{x},\epsilon_{\mathbf{x}})$ is disjoint from $B_{j} (j\geq i+2)$, then $C(\mathbf{x},\epsilon_{\mathbf{x}})$ is disjoint from $B_{j+1}$, but how to prove/disprove it?