In response to this question Can the Interval be Covered by Disjoint Cantor Sets? it was pointed out that the answer is, Yes: see Theorem 1.14 of Paul Bankston and Richard J. McGovern, Topological partitions, General Topology and its Applications 10 (1979), 215–229. http://www.mscs.mu.edu/~paulb/Paper/toppart.pdf. Constructions seem to involve the Axiom of Choice (see the partial answer to this question: Partitioning a metric space into Cantor sets)
My question is: Let $I$ be the closed interval and $Q$ the quotient of $I$ by such a partition. Then $Q$ is a compact $T_1$ space. Can $Q$ be Hausdorff?
Let $h:I\to I^2$ be the Hilbert curve. In particular, this map $h$ has the following property: for each dyadic interval $J\subset I$ (i.e., an interval of the form $[k/2^n,(k+1)/2^n]$), the image $h(J)$ is a product of two dyadic intervals. Let $f:I\to I$ be the first coordinate of $h$. Then $f$ is surjective and continuous, and thus a quotient map since $I$ is compact Hausdorff. Also, $f$ is nowhere locally constant, and for each $t\in I$, every neighborhood of $t$ contains uncountably many points $t'$ such that $f(t)=f(t')$ (since it contains some dyadic interval containing $t$ and then the image of that dyadic interval under $h$ contains an entire vertical line segment containing $h(t)$). Thus every fiber of $f$ is a Cantor set, so $f$ is a quotient map for a partition of $I$ into Cantor sets.