I'm studying Moretti's introduction to Spectral Theory and Quantum Mechanics in which he makes the following claim: (as far as I understand, I have reworded this to avoid discussing the irrelevant parts)

Suppose $M$ is an infinite set. Then $\text{card } M = \text{card } M \times \mathbb{N}$.

Is this true? I realized that this is equivalent to stating that all infinite sets can be partitioned into a countable number of sets, whose partitions all have the same cardinality as the set itself. How can this be proven?

The claim above is easily verifiable on sets such as $\mathbb{N}$ (partition it into multiples of primes) or in $\mathbb{R}$ (partition it into reals in the interval $(n, n+1]$ for $n \in \mathbb{Z}$).

This is because then, we can easily generate a surjection from $M$ to $M \times \mathbb{N}$ by mapping the $i$th partition of $M$ (which has a cardinality of $M$) trivially to the elements in $M \times \mathbb{N}$ of the form $(., i)$.

A surjection from $M \times \mathbb{N}$ to $M$ is trivial. So by Schroder-Bernstein they must have the same cardinality.

Possible idea: I think for any infinite set $M$ we know $\text{card } M = \text{card } M \times M$ by the axiom of choice. source

Since $M \times M$ can be easily partitioned into two parts each with cardinality $M$, then we now $M$ can be partitioned into two infinite sets with the same cardinality as M$.

We pick one of these two partitions and partition it again to two parts. We keep doing this and construct a countably infinite partition of sets with the cardinality of $M$.