A "Cantor set" is a topological space which is homeomorphic to the standard Cantor set $C$. In my answer to the question Another way for partition of perfect set by user 00GB I pointed out that the assertion "every (nonempty) perfect subset of $\mathbb R$ can be partitioned into $\mathfrak c$ Cantor sets" is an easy consequence of Theorem 1.14 of Paul Bankston and Richard J. McGovern, Topological partitions, General Topology and its Applications 10 (1979), 215–229 (pdf), which says that $\mathbb R$ itself can be partitioned into Cantor sets. The proof of Theorem 1.14 consists of first partitioning $\mathbb R$ into countably many Cantor sets and one set which is isomorphic to the space $\mathbb P$ of irrational numbers, and then using the fact that $\mathbb P$ can be partitioned into $\mathfrak c$ Cantor sets. This is very nice, but does not seem to generalize in any obvious way to higher-dimensional spaces.
Bankston & McGovern prove more general results (Theorem 1.12) about partitionability into Cantor sets, but only under special set-theoretic axioms; namely, they use Martin's axiom to prove that every complete separable metric space with no isolated points can be partitioned into Cantor sets, and they use the continuum hypothesis to prove the same for complete metric spaces of cardinality $\mathfrak c$ with no isolated points.
Question. Can those more general results of Bankston & McGovern be proved in ZFC? Can you prove in ZFC that every complete separable metric space with no isolated points, or even every complete metric space of cardinality $\mathfrak c$ with no isolated points, can be partitioned into Cantor sets?
By the way, since the Cantor set $C$ is homeomorphic to $C\times C$, any Cantor set can be partitioned into $\mathfrak c$ Cantor sets. Consequently, if a nonempty separable metric space $X$ can be partitioned into Cantor sets, then $X$ can be partitioned into $\mathfrak c$ Cantor sets.
Partial answer. At least the weaker result (for separable spaces) is provable in ZFC, if I haven't made a mistake. Here is a slightly more general statement:
Theorem. If $X$ is a nonempty separable metric space with no isolated points, and if $X$ has a dense subspace $Y$ which is completely metrizable, and if $|S|\lt\mathfrak c$, then $X\setminus S$ can be partitioned into $\mathfrak c$ Cantor sets.
Proof. Since $Y$ is a Polish space with no isolated points, every nonempty open subset (in the relative topology) of $Y$ contains a Cantor set, in fact, contains $\mathfrak c$ disjoint Cantor sets. Since $Y$ is separable, at most countably many of those disjoint Cantor sets can have nonempty interiors, and a closed set with empty interior is nowhere dense; so every nonempty open subset of $Y$ contains a Cantor set which is nowhere dense in $Y$. Let $U_0,U_1,U_2,\dots$ be a countable base for $Y$. We can choose pairwise disjoint nowhere dense Cantor sets $A_0,A_1,A_2,\dots$ with $A_n\subseteq U_n$. If we partition each $A_n$ into $\mathfrak c$ Cantor sets, we get a collection $\mathcal B$ of pairwise disjoint Cantor sets such that every nonempty open subset of $Y$ contains $\mathfrak c$ elements of $\mathcal B$. Since $Y$ is dense in $X$, every nonempty open subset of $X$ contains $\mathfrak c$ elements of $\mathcal B$. Moreover, we can assume that each element of $\mathcal B$ is disjoint from $S$.
Fix a well-ordering of $X\setminus S$ in which each point has $\lt\mathfrak c$ predecessors. We shall recursively construct new Cantor sets $C_x$ indexed by some points $x\in X\setminus S$; each $C_x$ will be the union of $\{x\}$ and countably many elements of $\mathcal B$.
At a certain stage of the construction, $\lt\mathfrak c$ Cantor sets $C_y$ have been constructed, and $x$ is the least point in $X\setminus S$ (with respect to the well-ordering) which is not covered by those sets $C_y$. Choose a sequence of disjoint open sets $V_n$ converging to $x$, that is, each neighborhood of $x$ contains all but finitely many of the $V_n$. For each $n$ choose a Cantor set $B_n\in\{B\in\mathcal B:B\subseteq V_n\}$ which is disjoint from all of the sets $C_y$ already constructed. Finally, let $C_x=\{x\}\cup B_0\cup B_1\cup B_2\cup\cdots$.
We have partitioned $X\setminus S$ into Cantor sets $C_x$.
Corollary. A nonempty Polish space with no isolated points can be partitioned into $\mathfrak c$ Cantor sets.