Prove that $\overline{W^j} \subset V^j$. I'm trying to prove the following Theorem from Trèves book:
Theorem 16.2: Let $\{V^j\}$ be a locally finite open covering of $\Omega$. To each $j$ there is an open subset $W^j$ of $\Omega$ such that $\overline{W^j} \subset V^j$ and when $j$ varies, the $W^j$ form an open covering (necessarily locally finite) of $\Omega$.
However I cannot conclude the final inclusion ($\overline{W^j} \subset V^j$).
Proof: We represent $\Omega$ as the union of a sequence of relatively compact open sets $\Omega_k$ such that $\overline{\Omega}_{k-1} \subset \Omega_k$ $(k\geq 1)$. For each $k$, let us consider the (finite) family of sets $V^j$, $j \in J$, which do intersect $\Omega_k$; let $J_k$ be the set of indices $j$ such that $V^j \cap \Omega_k \neq \emptyset$. Of course we have $\overline{\Omega}_k \subset \bigcup_{j \in J_k} V^j$. Consider then the function $d_k:\mathbb{R}^n \to [0,\infty)$ given by $$d_k(x)=\sup_{j \in J_k} \operatorname{dist}(x, \complement V^j)=\operatorname{dist}(x,\mathbb{R}^n\setminus V^j)\tag{*},$$ which is continuous in view of the fact that the supremum of a finite number of nonnegative continuous functions is continuous.
If $x \in \overline{\Omega}_k$, then $x \in V^j$ for some $j \in J_k$ and thus $\operatorname{dist}(x, \mathbb{R}^n\setminus V^j)>0$. We thus have $d_k(x)>0$ on $\overline{\Omega}_k$. Since $\overline{\Omega}_k$ is compact there exists $c_k=\inf_{x \in \overline{\Omega}_k}d_k(x)>0$. Thus, $(*)$ is $\geq c_k$ everywhere in $\overline{\Omega}_k$. For each $j \in J_k$ let us set
$$W_{k}^{j}=\{x \in V^j: \operatorname{dist}(x, \mathbb{R}^{n}\setminus V^j)>c_k/2\}.$$
It might happen that $W_{k}^{j}$ is empty. We moreover set $W_{k}^{j}=\emptyset$ for $j \in J \setminus J_k$. We set
$$W^j=\bigcup_{k=0}^{\infty}W_{k}^{j}.$$
Trèves says: It is easily seen that the open sets $W^j$ satisfy the requeriments in Theorem 16.2.

My question: How to prove that $\overline{W^j} \subset V^j$?

 A: I'm afraid that's another mistake in Trèves' book. With that construction it can happen (it's not actually rare) that $W^j = V^j$. That happens whenever $\inf c_k = 0$ because then for each $x \in V^j$ there are $k$ with $c_k/2 < \operatorname{dist}(x , \complement V^j)$, hence $x \in W^j_k$.
For a concrete example, let $\Omega = (0,1) \subset \mathbb{R}$, $\Omega_k = (2^{-2-k}, 1 - 2^{-2-k})$ for $k \in \mathbb{N}$, and for $j \in \mathbb{N}$ let
$$V^j = \biggl(1 - \frac{1}{2^{j}}, 1 - \frac{1}{2^{j+2}}\biggr)\,.$$
The point $x_k = 1 - 2^{-2-k}$ lies only in $V_{k+1}$, its distance to the complement is $2^{-3-k}$ and therefore $c_k \leqslant 2^{-3-k} \to 0$, which with the given construction gives $W^j = V^j$ for all $j$, and since no $V^j$ is closed in $\Omega$, $\overline{W^j} \not\subset V^j$ for all $j$.
A short fix is to modify the construction to use
$$W_k^j = \{ x \in V^j : \operatorname{dist}(x, \complement V^j) > c_k/2\} \setminus \overline{\Omega_{k-2}}$$
with $\Omega_n = \varnothing$ for $n < 0$.
Then $\{W^j_k : j \in J_k\}$ is an open cover of $\Omega_k \setminus \overline{\Omega_{k-2}} \supset \Omega_k \setminus \Omega_{k-1}$, whence $\{W^j : j \in J\}$ is an open cover of $\Omega$, and $\overline{W_k^j} \subset V^j$ for each $j$. Since we remove the points of $\overline{\Omega_{k-2}}$, for every $j \in J$ the family $\{W_k^j : k \in \mathbb{N}\}$ is locally finite, hence
$$\overline{W^j} = \overline{\bigcup_{k = 0}^{\infty} W_k^j} = \bigcup_{k = 0}^{\infty} \overline{W_k^j} \subset V^j\,.$$
