For a partition of $\mathbb{R}$ into countably infinite sets, must there be an almost-disjoint family of $2^{\frak c}$ many selectors? My question arises from a construction I gave in my recent
answer to a question of Alexander Pruss concerning large families of independent non-measurable sets of reals. In that argument, using
the continuum hypothesis and the existence of a thick Kurepa tree
$T$, I produced a family of $2^{\frak c}$ many Vitali sets $\{\
A_s\mid s\in[T]\ \}$, which was almost disjoint in the sense that
$A_s\cap A_t$ was countable whenever $s\neq t$. The only aspect of the Vitali relation that was used in the
construction was that the Vitali equivalence classes (equivalence
under rational translation) are countably infinite. Thus, the construction
proves:
Theorem. If the CH holds and there is a thick Kurepa tree,
then for every partition of $\mathbb{R}$ into countably infinite sets,
there is an almost-disjoint family of selectors of size $2^{\frak
c}$.
By almost-disjoint here, I mean that any two distinct elements
of the family have countable intersection; by selector, I
mean that each set in the family has exactly one element from each
equivalence class; and by a partition into countably infinite sets, I mean that we have an equivalence relation on $\mathbb{R}$ with every equivalence class countably infinite. To prove the theorem, simply label the nodes on the $\alpha^{th}$ level of $T$ with distinct members of the $\alpha^{th}$ equivalence class in the partition. Being thick, the tree has $2^{\frak c}$ many branches, each of which provides a selector, and any two such selectors can agree only up to some countable height in the tree, where those branches separate.
My question is whether I really needed those set-theoretic assumptions in order to make the conclusion.
Question. How much can one weaken the hypotheses of the theorem and still prove the conclusion? 
For example, can we drop the thick Kurepa tree assumption? Can we omit CH? Can we prove it in ZFC? Can one show the consistency with ZFC of a
counterexample?
 A: If CC (Chang's Conjecture) holds, then there are no $\aleph_2$ pairwise almost different $\omega_1\to\omega$ functions. For this, assume that $\{f_\alpha:\alpha<\omega_2\}$ are as described. By CC there is an elementary submodel $N$ such that $|N|=\aleph_1$ and 
$\delta=N\cap\omega_1<\omega_1$. (Technically, we have to consider the function $F(\alpha,\xi)=f_\alpha(\xi)$ for $\alpha<\omega_2$, $\xi<\omega_1$ to make the model of countable length. Then, apply CC to $M=(\omega_2;F,\omega_1,\in,\dots)$.) If $\alpha\neq \beta$ are in $N$, then they differ from some $\gamma<\omega_1$ on, by elementarity, there is such a $\gamma<\delta$, i.e., any two functions  in $\{f_\alpha:\alpha\in N\}$ differ from some $\gamma<\delta$. Specifically, $\{f_\alpha(\delta):\alpha\in N\}$ are distinct, a contradiction.
A: The existence of such a family of almost-disjoint selectors which is merely uncountable is equivalent to CH.  Indeed, suppose we have such a family; we can shrink it to a family of size $\aleph_1$.  Then for each equivalence class, there must be a pair of selectors that makes the same choice in that class, but each pair of selectors can only make the same choice countably many times.  Since there are only $\aleph_1$ pairs of selectors, there can only be $\aleph_1$ equivalence classes, and so $|\mathbb{R}|=\aleph_1$.  Conversely, assuming CH, it easy easy to construct such an uncountable family: identify the collection of equivalence classes with $\omega_1$, and order the $\alpha$th equivalence class with order type $\alpha$.  For each $\beta<\omega_1$, let the $\beta$th selector choose the $\beta$th element of each equivalence class (and an arbitrary element of the $\alpha$th equivalence class for $\alpha\leq\beta$).
