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

Theorem
Let $I = \left[{0 \,.\,.\, 1}\right]$ be the unit interval.

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

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

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

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

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

Then $J$ is of the second category, 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 $L \cap B = \varnothing$.

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

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

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 $u \in I \setminus C_j$, say $u \in C_m$.

Then if $U \cap B \cap \partial C_m = \varnothing$ 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 of the second category, since it is a closed subset of $I$.

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

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

But this is impossible, since if $U \cap C_k \ne \varnothing$ then $U \cap \left({B \setminus \partial C_k}\right) \ne \varnothing$.

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