Call a non-well-orderable set "simple'' if whenever it is partitioned into two pieces, one of them is well-orderable.
Does ZF prove that every non-well-orderable set has a simple non-well-orderable subset?
Partition of non-well-orderable sets
axiom-of-choiceset-theory
Related Solutions
This is a delicate matter. Do you mean all the well orders or just up to isomorphism?
For the former observe, for example, that if a set $A$ has one well-ordering then any permutation induces a different well-ordering, although of the same order type. For the natural numbers there are $2^{\aleph_0}$ many permutations so there are at least continuum many well-orders, and that is just of one isomorphism type! On the other hand, there can only be continuum many relations, so we have exhausted the cardinality.
We can continue by induction on the $\aleph$-cardinals, in fact, this is the only way we can resume. Why? Well, if a set has any well-ordering then it has to be in bijection with an ordinal, and if it is infinite this ordinal can be an $\aleph$-number. The argument for $\omega_1$, $\omega_2$ and so on is the same as above and this would require no choice at all.
If a set cannot be well-ordered, well... it has no well-ordering! However if we assume the axiom of choice then every set can be well-ordered and going through cardinals is enough. So arguing this claim for all sets is to require the full axiom of choice.
On the other hand, if you are interested in order types rather than mere orders than the claim that there are $2^A$ many is in fact to assert the Generalized Continuum Hypothesis, since for a set of cardinality $\kappa$ there are only $\kappa^+$ many ordinals of cardinality $\kappa$, and therefore only $\kappa^+$ many order types.
Whether or not $2^\kappa=\kappa^+$ is undecided in ZFC.
Proof that non-well-orderable sets are not equinumerous to their own cardinals (using Scott’s trick)
Assume that $|x|$ is not an ordinal. Then the element of $|x|$ is not well-orderable, so $x$ also is. By the definition of $|x|$, every $y\in |x|$ has the same rank, let it call $\alpha$. Then $|x|\subseteq R(\alpha+1)$, and we may assume $x\subseteq R(\alpha)$.
Let $f:R(\alpha)\times R(\alpha)\to R(\alpha)$ be a bijection. Observe that for each $y\in |x|$, $y$ is a subset of $R(\alpha)$. Now consider $f^"[y\times \{a\}]\subseteq R(\alpha)$ for each $a\in R(\alpha)$. Then we have $|R(\alpha)|$ pairwise disjoint copies of elements of $|x|$. Hence $|R(\alpha)|\le |\,|x|\,|$. Moreover, $x\subseteq R(\alpha)$ proves $|x|\le |R(\alpha)|$.
If $|x|<|R(\alpha)|$, then obviously $|x|<|\,|x|\,|$. Now assume that $|x|=|R(\alpha)|$. Without loss of generality, assume that $x=R(\alpha)$. I claim that $R(\alpha+1)$ has $|R(\alpha+1)|=2^{|R(\alpha)|}$ copies of $R(\alpha)$. In fact, we can prove
Claim If $|X|=|X|^2$, then $\mathcal{P}(X)$ has $2^{|X|}$ copies of $X$.
Proof. Let $f: X\times X\to X$ be a bijection. Then we have a family of pairwise disjoint sets $X_a:=f^"[\{a\}\times X]$ and a family of bijections $f(a,-): X\to X_a$.
For each $A\subseteq X$, consider $$Y_A = \bigcup_{a\in A}X_a.$$ Then $|X|\le |Y_A|= |A|\times |X|\le |X|^2=|X|$. (Proving $|Y_A|= |A|\times |X|$ requires the uniform family of bijections.) Since $A\neq B$ implies $Y_A\neq Y_B$, we are done.
It proves $|\,|R(\alpha)|\,|=2^{|R(\alpha)|}$.
Best Answer
Let me add another example, namely Cohen's first model. In that model we have a subset of $\Bbb R$ which is an infinite Dedekind-finite set, $A$. I claim that this $A$, and indeed any infinite Dedekind-finite subset of $\Bbb R$, is not well-orderable, and it is not simple.
To see that, first note that $A$ is not amorphous. Recall that a set is called amorphous if every partition into two parts has one of those be finite. But amorphous sets cannot be linearly ordered, so $A$ is not amorphous. Next, since $A$ is Dedekind-finite, every well-orderable subset is finite.
So, if a Dedekind-finite is not amorphous, it is not simple. Moreover, any subset of a Dedekind-finite set is Dedekind-finite. And so, once it can be linearly ordered, it is immediately hereditarily "not simple". And that is indeed the case in Cohen's model.