Can every infinite set be divided into pairwise disjoint subsets of size $n\in\mathbb{N}$? 
Let $S$ be an infinite set and $n$ be a natural number.  Does there exist partition of $S$ in which each subset has size $n$?



*

*This is pretty easy to do for countable sets.  Is it true for uncountable sets?

*If (1) is true, can it be proved without choice?

If you can write $S$ as a union of $n$ disjoint subsets, say $S_1 \cup \dots \cup S_n$ where the $S_i$'s all have the same cardinality, then take a bijection $\pi_i : S_1 \to S_i$ and consider the sets $\{  \{ \pi_i(s) \, | \, 1 \le i \le n \} \, | \, s \in S_1 \}$ which form a partition of $S$ with $n$ elements for each element of your partition. So we've reduced the problem to finding a partition of $S$ in $n$ subsets of the same cardinality... 
If $S$ has the cardinality of $\mathbb R$ (not a set theorist here... $c$ stands for $|\mathbb R|$?), we might as well just solve the problem for $\mathbb R$ using the bijection given from the bijection from Cantor-Bernstein-Schroeder's theorem (which doesn't rely on choice, apparently!). Take 
$$
S_1 = ]-\infty, 1], \quad \dots \quad S_i = \, ]i-1,i], \quad \dots \quad S_n = \, ]n-1,\infty[
$$
and in this case since you don't need choice to compute a bijection between those intervals, you're good to go.
Hope that helps,
