Question about nowhere dense condition in Baire category theorem application I'm asking a question in reference to Noah Schweber's answer in this question can open set in $\mathbb{R}^n$ be written as the union of countable disjoint closed sets?, which is itself a reference to GEdgar's answer here: Does not exist cover of $\mathbb{R}^n$ by disjoint closed balls. I'm having trouble with the second condition: showing that each $\partial C_{n}$ is a nowhere dense subset of $B$. 
First, a brief summary of the problem: show that no open subset $U \subseteq \mathbb{R}^{n}$ can be presented as a countable union of proper, closed and disjoint subsets of $\mathbb{R}^n$.
Assume for the sake of contradiction that $U = \bigcup_{n\geq1} C_{n}$, $C_{n} \subset \mathbb{R}^n$ closed and disjoint. Denote $B=\bigcup_{n\geq1}\partial C_{n}$. Then $B$ is complete, because it's closed: let $b \in U$ be a limit point of $B$, and $\lim_{n \to \infty} b_{n} = b$, $b_{n} \in B$. Then either $b \in B$ or $b \in C_{i}^{\circ}$ for some $i$. If the latter were to hold, i.e. $B(b, \varepsilon) \subseteq C_{i}$, we'd get a contradiction: since $b$ is the limit of $b_{n}$, we can find a $b_{n} \in \partial C_{k}$ such that $d(b_{n}, b)<\frac{\varepsilon}{2}$, and so if $k=i$, we can find a point $c$ from $C_{i}^{\complement}$ such that $d(b_{n}, c)<\frac{\varepsilon}{2}$, and so $d(b, c)<\varepsilon$, which is a contradiction. On the other hand, if $k \neq i$, then we can find a $c \in C_{k} \subseteq C_{i}^{\complement}$ such that $d(b_{n}, c) < \frac{\varepsilon}{2}$, which is again a contradiction. So $B$ is complete.
Here is my attempt at showing that each $\partial C_{n}$ is nowhere dense in $B$: since the boundary of a set is always closed, it suffices to show that $(\partial C_{n})^{\circ} = \emptyset$. If there were a $p \in (\partial C_{n})^{\circ}$, then $B(p, \varepsilon) \subseteq (\partial C_{n})^{\circ}$ for some $\varepsilon$, and since $C_{n}$ is closed, it contains its boundary, so $B(p, \varepsilon) \subseteq C_{n}$, so $p \in C_{i}^{\circ}$, which is a contradiction with $p \in \partial C_{n}$.
However, I've only proven that $\partial C_{n}$ is nowhere dense in $U$, and not in $B$. I'm trying to tweak my argument to make a contradiction by proving both $p \in (C_{n} \cap B)^{\circ}$ and $p \in \partial(C_{n} \cap B)$, but to no avail. Is it possible to just tweak the argument, or do I need a new one?
Since the question is 5 years old, and since my question is far too long for a comment, I've decided to open a new question instead of posting in the comments. If this does not merit a new question, I will gladly try to condense my argument/question into a comment in the original question.
 A: First let me note that $B$ would not usually be complete in the standard metric. Typically, $B$ would have limit points in $\partial U$. However, since $B$ is a closed subspace of the metrisable space $U$, it's a $G_{\delta}$-set in the Baire space $U$, hence itself a Baire space.
To show that $\partial C_n$ has empty interior in $B$ (which, since the $\partial C_n$ are closed is equivalent to being nowhere dense), let $x\in \partial C_n$, and let $V$ be a small ball around $x$, so that $x \in V \subset U$. Since $x\in \partial C_n$, it follows that $W := V\setminus C_n \neq \varnothing$. Pick a $k$ such that $C_k \cap W \neq \varnothing$. Then
$$V = \underbrace{(V \cap \overset{\Large\circ}{C_k}) \mathbin{\dot{\cup}} (V \cap \partial C_k)}_{=V\cap C_k \neq \varnothing} \mathbin{\dot{\cup}} \underbrace{(V\setminus C_k)}_{\supset V\cap C_n \neq \varnothing}.$$
Since $V$ is connected, it follows that $V\cap \partial C_k \neq \varnothing$. But that means $V\cap B \not\subset \partial C_n$. Since the small balls form a neighbourhood base of $x$, it follows that $x$ is not an interior point of $\partial C_n$ relative to $B$.
And thus we have our contradiction that the Baire space $B$ is meagre in itself.
