A set $B\subseteq\mathbb R$ is a Bernstein set if $B$ contains no nonempty perfect set while having nonempty intersection with every nonempty perfect set; in other words, if neither $B$ nor $\mathbb R\setminus B$ contains a nonempty perfect set. The existence of Bernstein sets is an easy consequence of the axiom of choice (see below); the axiom of choice is essential here since Bernstein sets are not Lebesgue measurable.

Let $B$ be a Bernstein set and let $C$ be the Cantor set. Then $C\cap B$ and $C\setminus B$ are sets of Lebesgue measure zero, neither of which contains a nonempty perfect set, and it's clear that at least one of them is uncountable. In fact $|C\cap B|=|C\setminus B|=\mathfrak c=2^{\aleph_0}$, because $C$ can be partitioned into $\mathfrak c$ nonempty perfect sets (since $C$ is homeomorphic to $C\times C$).

Construction of Bernstein sets. Since there are $\mathfrak c$ nonempty perfect subsets of $\mathbb R$, and since each of them has cardinality $\mathfrak c$, we can enumerate them in a transfinite sequence $(P_\alpha:\alpha\lt\mathfrak c)$ and we can recursively choose $a_\alpha,b_\alpha\in P_\alpha\setminus\bigcup_{\xi\lt\alpha}\{a_\xi,b_\xi\}$ with $a_\alpha\ne b_\alpha$; then $\{a_\alpha:\alpha\lt\mathfrak c\}$ and $\{b_\alpha:\alpha\lt\mathfrak c\}$ will be Bernstein sets. In a similar way we can construct $\mathfrak c$ disjoint Bernstein sets.

A set $B\subseteq\mathbb R$ is a Bernstein set if $B$ contains no nonempty perfect set while having nonempty intersection with every nonempty perfect set; in other words, if neither $B$ nor $\mathbb R\setminus B$ contains a nonempty perfect set. The existence of Bernstein sets is an easy consequence of the axiom of choice (see below); the axiom of choice is essential here since Bernstein sets are not Lebesgue measurable.

Let $B$ be a Bernstein set and let $C$ be the Cantor set. Then $C\cap B$ and $C\setminus B$ are sets of Lebesgue measure zero, neither of which contains a nonempty perfect set, and it's clear that at least one of them is uncountable. In fact $C\cap B$ and $C\setminus B$ both have cardinality $\mathfrak c=2^{\aleph_0}$ because $C$ can be partitioned into $\mathfrak c$ nonempty perfect sets (since $C$ is homeomorphic to $C\times C$).

Construction of Bernstein sets. Since there are $\mathfrak c$ nonempty perfect subsets of $\mathbb R$, and since each of them has cardinality $\mathfrak c$, we can enumerate them in a transfinite sequence $(P_\alpha:\alpha\lt\mathfrak c)$ and we can recursively choose $a_\alpha,b_\alpha\in P_\alpha\setminus\bigcup_{\xi\lt\alpha}\{a_\xi,b_\xi\}$ with $a_\alpha\ne b_\alpha$; then $\{a_\alpha:\alpha\lt\mathfrak c\}$ and $\{b_\alpha:\alpha\lt\mathfrak c\}$ will be Bernstein sets. In a similar way we can construct $\mathfrak c$ disjoint Bernstein sets.