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 to the question as stated is no, as others have explained. However, if we relax the hypothesis from disjoint to non-overlapping, then the answer is yes.
Two intervals $I_1$ and $I_2$ are non-overlapping if $I_1^{\circ}\cap I_2^{\circ}=\emptyset$; that is, if their interiors are disjoint. If the intervals are closed and non-overlapping, then they intersect at most in their boundaries. For example, in $\mathbb{R}$, the intervals $\left[0,\frac{1}{2}\right]$ and $\left[\frac{1}{2},1\right]$ are non-overlapping, but clearly not disjoint as they share the point $\frac{1}{2}$.
