Is $[0,1]$ a countable disjoint union of closed sets? Can you express $[0,1]$ as a countable disjoint union of closed sets, other than the trivial way of doing this?

A partial solution:
As Sargera began, if such a cover exists, it must be countably infinite. Let $[0,1]=\displaystyle\bigcup_{n=1}^{\infty}[a_n, b_n] = \displaystyle\bigcup_{n=1}^{\infty}I_n$ By the axiom of choice, construct $(x_n)$ by $x_n \in [a_n, b_n].$ Since $[0,1]$ is compact, there exists $(x_{n_k})$ such that $x_{n_k} \to x\in [0,1].$ Then there exists unique $s\in \mathbb{N}$ such that $x\in [a_s, b_s].$ If this interval is non-degenerate then corresponding to $\varepsilon = \frac{1}{2}\min\{x-a_s, b_s-x\}$ there exists $m \in \mathbb{N}$ such that $|x_{n_k}-x|< \varepsilon$ for $k \geq m.$ In particular $x_{n_{m}} \in (a_s, b_s)\subset [a_s, b_s] = I_s$ and since $x_{n_{m}} \in I_{n_{m}}$ by construction, we obtain a contradiction to the fact that the $I_n$ are disjoint.
If $I_s$ is degenerate, I haven't thought of a workaround yet. Maybe someone could suggest one.
