Is there a maximum to the amount of disjoint non-measurable subsets of the unit interval with full outer measure?

In this article, J. Cichon shows that the real line (and hence the unit interval) can be partitioned into continuum many Bernstein subsets. Of course, any Bernstein set has full outer measure.

In 1917 Luzin and Sierpinski proved there exists continuum many pairwise disjoint subsets of the interval $[0,1]$ such that each of these subsets has outer Lebesgue measure $1.$

Nikolai N. Luzin and Waclaw Sierpinski, Sur une décomposition d'un intervalle en une infnité non dénombrable d'ensembles non mesurables [On a decomposition of an interval into nondenumerably many nonmeasurable sets], Comptes Rendus Académie des Sciences (Paris) 165 (1917), 422-424.

See Gallica site for C. R. Paris volumes or Internet Archive copy of volume 165

Although I've mentioned this paper in various internet groups several times over the past 10+ years, I don't think I've ever mentioned why I find it so fascinating. First, that such an amazingly strong result was published so early -- little more than a decade after non-measurable sets were known to exist. Second, that such a result is so rarely mentioned in texts that deal with measure theory, despite that fact that very little background is needed to state the result, which can also easily be done in a one-sentence footnote.

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 $< 2^{\aleph_0}$.