Each neighborhood of a point $x$ of $\operatorname{Bd}A$ intersects infinitely many of the cubes from the collection $\mathcal{C}$. James R. Munkres I am reading "Analysis on Manifolds" by James R. Munkres.
The author wrote the following remark after the author finished the proof of Lemma 16.2:

We remark that the local finiteness condition holds for each point $x$ of $A$, but it does not hold for a point $x$ of $\operatorname{Bd}A$. Each neighborhood of such a point necessarily intersects infinitely many of the cubes from the collection $\mathcal{C}$, as you can check.

I proved that each neighborhood of a point $x$ of $\operatorname{Bd}A$ necessarily intersects infinitely many of the cubes from the collection $\mathcal{C}$.
But I am not sure my proof is ok or not.
If my proof is not ok, then please tell me a proof.
If my proof is ok but it is not a good proof, then please tell me a better proof.
Lemma 16.2 and its proof in this book are here:


Lemma 16.2. Let $\mathcal{A}$ be a collection of open sets in $\mathbb{R}^n$; let $A$ be their union. Then there exists a countable collection $Q_1,Q_2,\dots$ of rectangles contained in $A$ such that:
(1) The sets $\operatorname{Int}Q_i$ cover $A$.
(2) Each $Q_i$ is contained in an element of $\mathcal{A}$.
(3) Each point of $A$ has a neighborhood that intersects only finitely many of the sets $Q_i$.
Proof. It is not difficult to find rectangles $Q_i$ satisfying (1) and (2). Choosing them so they also satisfy (3), the so-called "local finiteness condition," is more difficult.
Step 1. Let $D_1,D_2,\dots$ be a sequence of compact subsets of $A$ whose union is $A$, such that $D_i\subset\operatorname{Int}D_{i+1}$ for each $i$. For convenience in notation, let $D_i$ denote the empty set for $i\leq 0$. Then for each $i$, define $$B_i=D_i-\operatorname{Int}D_{i-1}.$$ The set $B_i$ is bounded, beign a subset of $D_i$; and it is closed, being the intersection of the closed sets $D_i$ and $\mathbb{R}^n-\operatorname{Int}D_{i-1}$. Thus $B_i$ is compact. Also, $B_i$ is disjoint from the closed set $D_{i-2}$, since $D_{i-2}\subset\operatorname{Int}D_{i-1}$. For each $x\in B_i$, we choose a closed cube $C_x$ centered at $x$ that is contained in $A$ and is disjoint from $D_{i-2}$; also choose $C_x$ small enough that it is contained in an element of the collection of open sets $\mathcal{A}$. SEe Figure 16.2.
The interiors of the cubes $C_x$ cover $B_i$; choose finitely many of these cubes whose interiors cover $B_i$; let $\mathcal{C}_i$ denote this finite collection of cubes. See Figure 16.3.
Step 2. Let $C$ be the collection $$\mathcal{C}=\mathcal{C}_1\cup\mathcal{C}_2\cup\dots;$$ then $\mathcal{C}$ is a countable collection of rectangles (in fact, of cubes). We show this collection satisfies the requirements of the lemma.
By construction, each element of $\mathcal{C}$ is a rectangle contained in an element of the collection $\mathcal{A}$. We show that the interiors of these rectangles cover $A$. Given $x\in A$, let $i$ be the smallest integer such that $x\in\operatorname{Int}D_i$. Then $x$ is an element of the set $B_i=D_i-\operatorname{Int}D_{i-1}$. Since the interiors of the cubes belonging to the collection $\mathcal{C}_i$ cover $B_i$, the point $x$ lies interior to one of these cubes.
Finally, we check the local finiteness condition. Given $x$, we have $x\in\operatorname{Int}D_i$ for some $i$. Each cube belonging to one of the collections $\mathcal{C}_{i+2},\mathcal{C}_{i+3},\dots$ is disjoint from $D_i$, by construction. Therefore the open set $\operatorname{Int}D_i$ can intersect only the cubes belonging to one of the collections $\mathcal{C}_1,\dots,\mathcal{C}_{i+1}$. Thus $x$ has a neighborhood that intersects only finitely many cubes from the collection $\mathcal{C}$.


We remark that the local finiteness condition holds for each point $x$ of $A$, but it does not hold for a point $x$ of $\operatorname{Bd}A$. Each neighborhood of such a point necessarily intersects infinitely many of the cubes from the collection $\mathcal{C}$.

Each neighborhood of a point $x$ of $\operatorname{Bd}A$ necessarily intersects infinitely many of the cubes from the collection $\mathcal{C}$.
The following is my proof of this fact:

Let $x_0\in\operatorname{Bd}A$.
Let $U$ be any neighborhood of $x_0$.
Let $d(x_0,D_i):=\min\left\{|x_0-d|\mid d\in D_i\right\}$ for $i\in\{1,2,\dots\}$.
Then, $d(x_0,D_i)>0$ for any $i\in\{1,2,\dots\}$ since $x_0\notin A$.
Assume that $\left\{i\in\{1,2,\dots\}\mid U\cap D_i\neq\emptyset\right\}$ is bounded above.
Let $n_0:=\max\left\{i\in\{1,2,\dots\}\mid U\cap D_i\neq\emptyset\right\}$.
Let $r$ be a positive real number such that $\left\{y\in\mathbb{R}^n\mid |y-x_0|<r\right\}\subset U$.
Then, $\left\{y\in\mathbb{R}^n\mid |y-x_0|<\min\{d(x_0,D_{n_0}),r\}\right\}\cap A\neq\emptyset$.
Let $y\in\left\{y\in\mathbb{R}^n\mid |y-x_0|<\min\{d(x_0,D_{n_0}),r\}\right\}\cap A$.
Since $y\in A$, there exists $m\in\{1,2,\dots\}$ such that $y\in D_m$.
Since $y\notin D_{n_0}$, $n_0<m$.
This is a contradiction.
So, $\left\{i\in\{1,2,\dots\}\mid U\cap D_i\neq\emptyset\right\}$ is not bounded above.
Therefore, each neighborhood of a point $x$ of $\operatorname{Bd}A$ necessarily intersects infinitely many of the cubes from the collection $\mathcal{C}$, as you can check.

 A: I think there exists a more convenient idea. (My following reasoning is not a rigorous proof)
We know that $A=\bigcup _{n}D_n$ where each $D_n$ is compact. Then we know that in $\mathbb R^n$, finitely many unions of the closed set are closed. Thus, if we don't let $A$ be an empty set, then $A$ must be a set obtained by infinitely many unions of compact sets $D_n$
Then consider a boundary point $x_0$ of $A$, since $A$ is open, so $x_0\notin A$ . Thus it is not an isolated point of $A$, and since it is the boundary point, it can only be a limit point of $A$.
Thus, for all open sets $U$ that contains $x_0$, $U\backslash \{x_0\}\cap A \ne \emptyset$.
That is $U\backslash \{x_0\}\cap (\bigcup_n D_n)\ne \emptyset$. Thus $U\backslash \{x_0\}$ must intersects with infinitely many $D_n$. Since if suppose it only intersects with finitely many $D_n$. Then just pick the largest $n=N$ such that $U\backslash \{x_0\}$ doesn't intersect with any $D_n$ with $n>N$, so $ U\backslash \{x_0\}$ doesn't intersect with $D_{N+1}$. This is a contradiction since $D_{N+1}\supset D_{N}$. Thus if $U\backslash \{x_0\}$ intersects with $D_N$ , it must intersect with $D_{N+1}$.
Now we know that $U\backslash \{x_0\}$ intersects with infinitely many $D_n$, we intuitively know that it intersects with infinitely many $B_i$. That is, it intersects with infinitely many $Q_i$
