I can solve this assuming the continuum hypothesis. (Edit: CH isn't needed, see below.) Lemma: if $A$ is a countable set and $(S_m)$ is a sequence of uncountable sets then we can find a sequence of disjoint countable sets $(T_n)$ such that $A \cap T_n = \emptyset$ for all $n$ and $S_m \cap T_n \neq \emptyset$ for all $m$ and $n$. [Proof: Choose a countable subset $S_1'$ of $S_1 \setminus A$, enumerate it, and put the $n$th element in $T_n$. Then choose a countable subset $S_2'$ of $S_2 \setminus (A \cup S_1')$, enumerate it, and put the $n$th element in $T_n$. Proceed in this way. Each $S_k'$ is countable so the difference $S_{k+1} \setminus (A \cup S_1' \cup \cdots \cup S_k')$ is always uncountable.]

Now there are $2^{\aleph_0}$ open subsets of $[0,1]$ because any open subset is a union of rational intervals and there are only countably many rational intervals. So there are only $2^{\aleph_0}$ closed subsets of $[0,1]$. Assume CH and enumerate the closed subsets of $[0,1]$ of positive measure as $C_\alpha$ for $\alpha < \aleph_1$. Observe that each $C_\alpha$ is uncountable. Now we construct disjoint countable sets $T_{\alpha,\beta}$ for $\beta < \alpha < \aleph_1$ by recursion on $\alpha$ as follows. At step $\alpha$ let $A = \bigcup_{\beta' < \alpha' <\alpha} T_{\alpha',\beta'}$ (all the $T$s constructed so far, a countable union of countable sets) and apply the lemma to this $A$ and the sets $C_\beta$ for $\beta < \alpha$. There are only countably many $\beta < \alpha$ so we can do this, and we can relabel the resulting sets $T_n$ as $T_{\alpha,\beta}$ for $\beta < \alpha$.

After this process is complete, for each $\beta$ let $T_\beta = \bigcup_{\alpha > \beta} T_{\alpha,\beta}$. Then the sets $T_\beta$ are disjoint, there are $\aleph_1$ of them, and each one intersects every closed subset of $[0,1]$ of positive measure, so each of them has full outer measure.

Edit: actually this doesn't need CH. Every closed set is the union of a countable set and a perfect set, so if it has positive measure then it contains $2^{\aleph_0}$ elements. That's enough to keep the induction going for $\alpha < 2^{\aleph_0}$ since each $T_{\alpha,\beta}$ will have cardinality $< 2^{\aleph_0}$.