Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets

Theorem
Let $\mathbb I = \closedint 0 1$ be the closed unit interval.

Then $\mathbb I$ cannot be expressed as the union of a countably infinite set of pairwise disjoint closed sets.

Proof
$\displaystyle \mathbb I = \bigcup_{i \mathop = 1}^\infty C_i$ where $\set {C_i}$ is a set of pairwise disjoint closed sets.

Let:
 * $\ds B = \bigcup \partial C_i = \mathbb I \setminus \bigcup C_i^\circ$

where $\partial C_i$ is the boundary of $C_i$ and $C_i^\circ$ is the interior of $C_i$.

Let $J \subseteq \mathbb I$ be a subinterval of $\mathbb I$.

Then $J$ is non-meager, and so some $C_k$ is dense in some open interval $L \subseteq J$.

Since $C_k$ is closed we have that $L \subseteq C_k^\circ$ and so $L \cap B = \O$.

So $J$ contains an open subset $L$ disjoint from $B$.

So $B$ is nowhere dense in $\mathbb I$.

So every open interval $U$ containing some $x \in \partial C_j$ must intersect $B \setminus \partial C_j$.

This is because $U$ is an open neighborhood of $x$ and therefore contains some point $u \in \mathbb I \setminus C_j$, say $u \in C_m$.

Then if $U \cap B \cap \partial C_m = \O$ it follows that $C_m^\circ \cap U$ is a non-empty open set in $U$. which is also relatively closed.

Now $B$ is itself non-meager, since it is a closed subset of $\mathbb I$.

Thus some $\partial C_k$ is dense in some non-empty open set $U \cap B$, for some open interval $U$ in $\mathbb I$.

Since $\partial C_k$ is closed, this means $\partial C_k \cap U = B \cap U$.

But this is impossible, since if $U \cap C_k \ne \O$ then $U \cap \paren {B \setminus \partial C_k} \ne \O$.

This contradiction shows that $\mathbb I$ cannot be expressed as $\ds \bigcup_{i \mathop = 1}^\infty C_i$.