Yes. You need some choice. The assertion "Every infinite set has a countable subset" is more than enough, and we can use even less (although people often go with more and assume axiom of countable choice).
A set is called Dedekind infinite when it has a proper subset with the same cardinality. That is $A$ is Dedekind infinite when there is some $B\subsetneq A$ such that $|A|=|B|$. In ZFC every infinite set is Dedekind infinite.
A set which is infinite but not Dedekind infinite is called infinite Dedekind finite set (commonly abbreviated to iDf or Dedekind finite). That is $A$ is iDf if for every $B\subsetneq A$ we have $|B|<|A|$. It does not mean that you cannot split an iDf into two infinite sets.
For example, take two disjoint iDf sets $A,B$ then $A\cup B$ is also iDf but can be split into $A$ and $B$.
Important classification of Dedekind infinite sets is as follows: $$A\text{ is Dedekind infinite}\iff\aleph_0\le|A|$$
We do not require that $A$ will be well orderable, but we do require it will have a countable subset. In particular this shows how assuming the above gives us that every infinite set can be split into two infinite subsets.
This still not enough to prove that for every infinite $A$ there are $B,C\subseteq A$ such that $|A|=|B|=|C|$ and $B\cap C=\emptyset$. However it was proved that the axiom of choice does not follow from the following fact:
For every infinite cardinal $\mathfrak p$ we have: $\mathfrak p = \mathfrak p+\mathfrak p$. (I am unfamiliar with the proof, announced by Sageev in 1973 and published a couple years later).
This means that you can have that every cardinal can be split into two equinumerous parts without the axiom of choice.
Lastly, $A$ is called amorphous if $A$ cannot be split into two infinite subsets. That is to say, $B\subseteq A$ then either $B$ finite or $A\setminus B$ finite. This is a stronger notion than that of infinite Dedekind finite. This is due to the fact that it may be possible to linearly order an infinite Dedekind finite set, while it is impossible to linearly order an amorphous set.
The proof for this fact is as follow: Suppose $<$ is a linear ordering of $A$, take the cut at $a\in A$ to be $\langle a\!\!\mid =\{b\in A\mid b<a\}$. Clearly $a\mapsto\langle a\!\!\mid$ is a bijection between $A$ and the set of cuts in $<$, therefore the set of cuts is also amorphous. Since every cut is a subset of $A$ it is either finite or co-finite, and every set of cuts is finite or co-finite (with respect to the set of all cuts).
If only finite many cuts are finite, then only finitely many cuts are co-finite, which is a contradiction to the fact there are infinitely many cuts and each is a subset of $A$. The same argument shows that there cannot be only finitely many infinite cuts.
In Cohen's first model showing that ZF is consistent with the negation of AC was given by adding an iDf set of reals, which can be linearly ordered (but not well ordered). In particular in this model every set can be linearly ordered (for a slightly more detailed survey, see my answer here).
A note on consistency strength:
From the assumption that ZF is consistent you have that ZFC is consistent, and by forcing you have that "ZF+There is an iDf set+Every set if linearly orderable" is consistent, and by a different forcing you have that "ZF+Amorphous" is consistent. The last two clearly proving the consistency of ZF (if indeed they are consistent).
However, if we measure the consistency by how much we contradict the axiom of choice... well, in this case just as "axiom of countable choice" is stronger than "Every infinite set is Dedekind infinite" (i.e. the former assertion proves the latter over ZF), we have that:
"There exists an amorphous set" proves "There exists an iDf set", while the opposite is not true, as witnessed by the consistency of "ZF+There exists iDf set+Every set can be linearly ordered", since the last assertion is inconsistent with amorphous sets but still consistent with iDf sets.
And by its definition an iDf set which is not amorphous can be written as the disjoint union of two infinite sets. Therefore assuming that there exists an iDf set, but there are no amorphous sets is enough to ensure that every infinite set splits into two infinite sets. Whether or not this implies any other forms of choice (multiple choice, finite choice, choice from pairs, choice from well orderable sets, choice from a well orderable collection of well ordered sets, etc etc.), I do not know the answer for that. I'd be happy to look into this question, but this may take a few days due to prior engagements I have.
Lastly, two papers which may be interesting to this topic of conversation:
J.K.Truss, Classes of Dedekind Finite Cardinals, Fundamenta Mathematicae 84, 187-208, 1974.
In which seven notions of Dedekind finite cardinals (five in addition to finite sets and the one mentioned in my answer, which is the "canonical type" of infinite Dedekind finite cardinals). The paper is not very technical (I think) and I think that one can understand most of the results given without deep background in forcing or permutation models.
J.K.Truss, The structure of amorphous sets, Annals of Pure and Applied Logic, 73 (1995), 191-233.
This paper deals with amorphous sets, it is longer and more difficult than the previous one.
(I had to dig quite deep into the internet to find these online, and I cannot recall where I had found both papers. Many, many many thanks to Theo for finding the links.)
Yes, no, and consequently also no.
The first one is actually a theorem: if there is an infinite Dedekind-finite set, then there is one which can be mapped onto a strictly larger set. That means that the surjection defines the partition you are looking for.
Suppose that $A$ is Dedekind-finite, then $S_{inj}(A)$, which is the set of all injective finite sequences from $A$, is also Dedekind-finite. For if it wasn't, then we would have a countably infinite set of enumerated finite sets, and their union is therefore a countably infinite subset of $A$, and that is impossible.
Now take, for example $S_{inj}(A)$ without the empty sequence. Then the function erasing the first coordinate from each non-empty sequence is surjective onto $S_{inj}(A)$. Indeed, taking any subset which contains "all sequences of length $n\in I$" for some fixed infinite $I\subseteq\omega$ will work in a similar way.
For the second question, note that if $A$ is amorphous and $f\colon A\to X$ is surjective, then $X$ must be amorphous (or finite, if you'd like to require that amorphous implies infinite). For otherwise, simply partition $X$ into two infinite subsets and look at their preimages, both would have to be infinite subsets of $A$ which are disjoint.
How does that help us? Well, if $X$ is infinite, then the preimage of each point is finite. Now, starting with some $x\in X\setminus A$, consider $f^{-1}(x)=A_0$, then $f^{-1}(A_0)=A_1$ is also finite, and it must be distinct from $A_0$, and then we can continue by recursion and define a countably family of finite subsets of $A$. But this is impossible as the power set of an amorphous set is itself Dedekind-finite.
Note, however, that if $X$ is amorphous, its partitions are either finite; almost all singletons; or incomparable with $X$ itself. So we can get a partition that is incomparable, just not a partition that is strictly bigger.
Best Answer
Well. There's not a whole lot we can say at this level of generality. For example, if $A$ is strongly amorphous, i.e. every partition of $A$ is almost entirely made of singletons, then it is easy to see that any permutation of $A$ can only move finitely many points, since it decomposes $A$ into orbits, and those must be finite.
On the other hand, if $A$ is a $2$-amorphous set, i.e. there is a partition into pairs, then there is also a permutation which simply switches the elements of each pair, and therefore has infinitely many nontrivial orbits.
On the third hand,1 if $A$ is amorphous, then any infinite partition of $A$ must also be amorphous, and so it is easy see that the behaviour across the different orbits of a given permutation must be the same almost everywhere.
Finally, and this is where it starts getting really weird, there's a lot of structure that is compatible with certain "level" of amorphous-ness2 sets. Given any first-order structure whose first-order definable subsets are finite or co-finite, and has sufficiently rich automorphism group (enough to witness strong homogeneity), we can arrange a model where this structure has an amorphous "version".
For example, an infinite dimensional vector space over a finite field. But also a set with some partitions that satisfy certain basic rules can be used.
So, if we have an infinite set with a partition into $3$-element sets, we can also add a function to this structure, which is a permutation that generates that partition; but we can also elect to not add such a function, and ensure that none exist, i.e. that a $3$-amorphous set will only have finitely supported permutations.
Therefore, at this "maximal" level of generality, of just talking about amorphous sets, there's not a whole lot that we can say. Sam Tarzi, a former student of Peter Cameron, wrote a Ph.D. thesis titled "Group Actions on Amorphous Sets and Reducts of Coloured Random Graphs" and they have a preprint (presumably derived from that work, although seems to have never been through the peer review process), titled "Group Actions On Amorphous Sets" that might be of interest to you.
Here's something that I couldn't find in the literature, but I took a few minutes to make the argument. But I am almost certain this should be known.
Theorem. If $A$ is amorphous, then $\operatorname{Sym}(A)$ is Dedekind-finite.
Proof. Suppose that $\{p_n\mid n<\omega\}$ is a sequence of distinct permutations of $A$. Recall that $[A]^{<\omega}$, the set of finite subsets of $A$, is Dedekind-finite and the each orbit of $p_n$ must be finite.
For a given $a\in A$, let $O(a)$ be the closure of $\{a\}$ under all the $p_n$s. We first claim that $O(a)$ must be finite. Since orbit of $a$ under each $p_n$ must be finite, there must be finitely many "types" of orbits, as $[A]^{<\omega}$ is Dedekind-finite; taking $O_0(a)$ as the union these orbits, we define $O_{n+1}(a)=\bigcup\{O(x)\mid x\in O_n(a)\}$, then each $O_{n+1}$ is finite, and this is an increasing sequence of sets in $[A]^{<\omega}$, so it must stabilise, and therefore $O(a)$ must be finite.
Next, as $\{O(a)\mid a\in A\}$ must form a partition of $A$, all but finitely have the same size. More importantly, though, all of the interesting properties that follow from how $p_n$, for whatever $n$, acts on $O(a)$ must also be the same for all but finitely many examples.
In particular, for a fixed $n$, all but finitely many $a$s have the same "type" of $p_n$ on $O(a)$. In other words, for a cofinite set of $a$ and $b$ there is a bijection mapping $O(a)$ to $O(b)$ which commutes with $p_n$.
So, letting $A_n$ be the [finite] set of "atypical $O(a)$", we have, by Dedekind-finiteness, that $\{A_n\mid n<\omega\}$ is just finitely many sets, repeating. Therefore $A^*=\bigcup A_n$ is a finite set, and for any $a,b\notin A^*$, $p_n$ has the same type on $O(a)$ and $O(b)$. But since each $O(a)$ is finite, there can only be finitely many of these types, so must be able to distinguish infinitely many $p_n$s just by examining $A^*$.
But now we ran into a bit of a problem! Since $A^*$ is finite, there can only be finitely many permutations of $A^*$, so we cannot distinguish infinitely many permutations at once!
So, the only way to resolve this is that there was no countably sequence of permutations, and therefore $\operatorname{Sym}(A)$ is Dedekind-finite as well. $\square$
Amorphous sets are weird, and apparently they have more than two hands. Don't get me started on the number of fingers each of those have.
Amorphity? amorphousity? amorphousnessicity? who knows...