Although [this problem](https://mathoverflow.net/questions/275633/existence-of-a-diagonal-set-in-certain-set-systems) got an accepted answer, I still feel there are more questions to ask in this direction.



> 
> Suppose that to every $\alpha\in\mathbb R$ there corresponds a subset $A\_\alpha\subseteq\mathbb R$ with $|A\_\alpha|=\mathfrak c$. Does there always exist a set $D\subseteq\mathbb R$ such that $D\cap A\_\alpha\ne\varnothing$ and $A\_\alpha\nsubseteq D$ for each $\alpha\in\mathbb R$? Equivalently, does there always exist a choice function $f\colon\mathbb R\to\mathbb R$ such that $f(\alpha)\in A\_\alpha\nsubseteq\cup\_{\alpha\in\mathbb R}\{f(\alpha)\}$ for each $\alpha\in\mathbb R$? Equivalently, do there always exist *two* choice functions $f,g\colon\mathbb R\to\mathbb R$ such that $f(\alpha),g(\alpha)\in A\_\alpha$ for all $\alpha\in\mathbb R$, while $\{f(\alpha)\}\_{\alpha\in\mathbb R}$ and $\{g(\alpha)\}\_{\alpha\in\mathbb R}$ are disjoint?
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As [domotorp](https://mathoverflow.net/users/955/domotorp) obsrved in the comments, I am in fact asking whether any hypergraph with the vertex set, the number of edges, and the size of every edge all being of cardinality $\mathfrak c$, is $2$-colorable.



Choosing *two* representatives
I assume by $\aleph$ you [mean](https://math.stackexchange.com/questions/444273/how-many-infinite-cardinals-are-smaller-than-aleph?noredirect=1#comment4871997_444273) $\mathfrak c$, the cardinality of the continuum. You can build $D$ by transfinite recursion: Well-order the continuum in type $\mathfrak c$. At stage $\alpha$ you add a point of $A\_\alpha$ to your set, and one to its complement. You can always do this because at each stage fewer than $\mathfrak c$ many points have been selected.


The argument is standard, by the way, typically used to build Bernstein sets, that is, sets such that neither it nor its complement contains a perfect subset. The point is that there are precisely $\mathfrak c$ many perfect subsets of $\mathbb R$ (and any perfect set has size $\mathfrak c$).


Note that this particular case also shows that some amount of choice is needed to ensure the existence of $D$. This is because (modulo the consistency of an inaccessible cardinal) it is [consistent](http://www.ams.org/mathscinet-getitem?mr=265151) with $\mathsf{ZF}+\mathsf{DC}$ that every uncountable set of reals contains a perfect subset. ($\mathsf{DC}$ is the axiom of dependent choices, which suffices to develop classical analysis.) 


