Proof that non-well-orderable sets are not equinumerous to their own cardinals (using Scott's trick) Azriel Levy's Basic Set Theory (the 2002 Dover edition) contains the following problem (Chapter 3, Exercise 2.24):

Prove that $|x|$ is an ordinal iff $|x| \approx x$, (i.e., iff $|\,|x|\,| = |x|$). If $\mathfrak{a}$ is not a well-ordered cardinal then $|\mathfrak{a}| > \mathfrak{a}$.
Hint. Use the fact that for infinite $\alpha$, $R(\alpha) \times 2 \approx R(\alpha)$, thus $R(\alpha)$ includes two disjoint
copies of itself.

$R(\alpha)$ is defined as the set of sets whose ranks are strictly less than $\alpha$.
The book uses the following definition of cardinal numbers:

The cardinal of $x$, or, synonymously, the cardinality of $x$, which we
shall denote by $|x|$, is
(a) the least ordinal $\alpha$ equinumerous to $x$, if $x$ is well-orderable, and
(b) the set of all sets $y$ of least rank which is equinumerous to $x$, otherwise.

In other words, it uses representative ordinals for well-orderable sets, and Scott's trick for non-well-orderable ones.
The part of this problem I am specifically having trouble with is proving that if $|x|$ is not an ordinal, then $|x| \not\approx x$. The two proof techniques that I can think of which seem relevant are some kind of diagonalization argument, or assuming that $|x| \approx x$ and using the relevant bijection to obtain a well-ordering of $x$. However, after spending several hours thinking about this problem, I cannot come up with a successful way of doing either.
I'm pretty sure that the hint is not relevant here (I found it useful in proving the second sentence of the problem, given the truth of the first, so I assume that's the part it was intended for).
This doesn't seem like a problem that is intended to be this difficult, so I think have probably overlooked something. Any help solving it would be greatly appreciated.
 A: 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 $|\,|R(\alpha)|\,|=2^{|R(\alpha)|}$.
