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?
The answer is no. In fact, as Steve D said, we have a theorem that holds for a wide class of spaces, which includes closed intervals, circles, balls and cubes. It was proved by Sierpi&#324;ski in $1918$ $[1]$. You can find the proof in the book "General Topology" by Ryszard Engelking, but I'll post here since it's not easy to find it online. First a definition: a topological space is called a continuum if it is a compact connected Hausdorff space. The precise statement is the following:

Theorem (Sierpi&#324;ski). If a continuum $X$ has a countable cover $\{X_i\}_{i=1}^{\infty}$ by pairwise disjoint closed subsets, then at most one of the sets $X_i$ is non-empty.

In order to prove this we'll need the following lemmas:

Lemma $1$. Let $X$ be a continuum. If $F$ is a non-trivial closed subset of $X$, then for every component $C$ of $F$ we have that $\text{Bd}(F) \cap C$ is non-empty.

Proof. Let $x_0$ be in $C$. Since $X$ is Hausdorff compact, quasicomponents coincide with components, so $C$ is the intersection of all open-closed sets in $F$ which contain $x_0$. Suppose that $C$ is disjoint from $\text{Bd}(F)$. Then, by compactness of $\text{Bd}(F)$, there is one open-closed set $A$ in $F$ containing $x_0$ and disjoint from $\text{Bd}(F)$. Take an open set $U$ such that $A = U \cap F$. Thus the equality $A \cap \text{Bd}(F) = \emptyset$ implies that $A = U \cap \text{Int}(F)$, so $A$ is open in $X$. But $A$ is also closed in $X$, and contains $x_0$, so $A=X$. But then $\text{Bd}(F) = \emptyset$, which is not possible since $F$ would be non-trivial open-closed in $X$. $\bullet$

Lemma $2$. If a continuum $X$ is covered by pairwise disjoint closed sets $X_1, X_2, \ldots$ of which at least two are non-empty, then for every $i$ there exists a continuum $C \subseteq X$ such that $ C \cap X_i = \emptyset$ and at least two sets in the sequence $C \cap X_1, C \cap X_2, \ldots$ are non-empty.

Proof. If $X_i$ is empty then we can take $C = X$; thus we can assume that $X_i$ is non-empty. Take $j \ne i$ such that $X_j \ne \emptyset$. Since $X$ is Hausdorff compact, there are disjoint open sets $U,V \subseteq X$ satisfying $X_i \subseteq U$ and $X_j \subseteq V$. Let $x$ be a point of $X_j$ and $C$ the component of $x$ in the subspace $\overline{V}$. Clearly, $C$ is a continuum, $ C \cap X_i = \emptyset$ and $ C \cap X_j \ne \emptyset$. By the previous lemma, $C \cap \text{Bd}( \overline{V}) \ne \emptyset$ and since $X_j \subseteq \text{Int}(\overline{V})$, there exist a $k \ne j$ such that $C \cap X_k \ne \emptyset$. $\bullet$

Now we can prove the theorem:

Proof. Assume that at least two of the sets $X_i$ are non-empty. From lemma $2$ it follows that there exists a decreasing sequence $C_1 \supseteq C_2 \ \supseteq \ldots$ of continua contained in $X$ such that $C_i \cap X_i = \emptyset$ and $C_i \ne \emptyset$ for $i=1,2, \ldots$ The first part implies that $\bigcap_{i=1}^{\infty} C_i = \emptyset$ and from the second part and compactness of $X$ it follows that $\bigcap_{i=1}^{\infty} C_i \ne \emptyset$. $\bullet$

The Hausdorff hypothesis is fundamental. For example, consider $X$ a countable infinite set with the cofinite topology. Then $X$ is compact, connected and a $T_1$-space. However, we can write $X$ as a disjoint union of countable singletons, which are closed.

$[1]$ Sierpi&#324;ski, W: Un théorème sur les continus, T&ocirc;hoku Math. J. 13 (1918), 300&ndash;305.
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 $|x_{n_k}-x|&lt; \varepsilon$ 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.