One way to prove this is to first show that $\kappa +\mu =\max\{\kappa ,\mu \}$ when either $κ$ or $μ$ are infinite cardinals. This is assumed in the proof below.

I searched the web for this approach and found it here - theorem B3 in the appendix combines both, showing first that $\kappa +\mu =\max\{\kappa ,\mu \}$ and then that $\kappa \times \mu =\max\{\kappa ,\mu \}$.

We begin with a lemma.

Lemma 1: Let $B$ be a subset of an infinite set $A$ and $f: B \to B \times B$ a surjective function. Then $|B| \le |B \times B| \le |B| \le |A|$. Moreover, if $|B|$ is indeed less that $|A|$, then $f$ can be extended to a surjective function $D \to D \times D$, with $B$ a proper subset of $D$.

Proof: For the first part, apply elementary cardinality theory. For the second part, the we can find an infinite set $U$ that is disjoint from $B$, so that $|U| = |B|$; we also have the identity

$\tag 1 (B \cup U) \times (B \cup U) = (B \times B) \cup (B \times U) \cup (U \times B) \cup (U \times U)$

a disjoint union of four pieces all having a cardinality of $|B|$.

The function $f$ takes care of the first piece, and a cardinality argument allows us to surjectively cover the remaining three pieces with a function operating on the set $U$ as a domain. So we can extend $f$ to $D = B \cup U$. $\quad \blacksquare$

We are now ready to prove the main result:

Proposition 2: For any infinite set $A$,

$\tag 2 | A \times A | = |A|$

Proof

We only have to show that $|A| \ge |A \times A|$.

Consider the collection of all $(B,\phi)$ where $B \subseteq A$ and $\phi : B \to B \times B$ is a surjection. This collection is nonempty since there is a surjection $ \mathbb N \to \mathbb N \times \mathbb N$.

This collection can be partially ordered by $(B,\phi) < (C,\psi)$ if $B \subseteq C$ and $\psi|_B = \phi$. Every chain has an upper bound; simply take the union of the graphs of the functions in the chain, defining a surjective function $D \to D \times D$.

By Zorn's lemma there is a maximal element $(\hat B,\hat \phi)$. By lemma 1, we can proceed under the assumption that $|B| \lt |A|$, since otherwise we can use $\hat \phi$ to establish (2). But then lemma 1 also provides a surjective extension of $\hat \phi$, contradicting that $(\hat B,\hat \phi)$ was a maximum element, i.e. no such extension can be found. $\quad \blacksquare$

This proof was arrived at by 'lifting' the proof that $|A \times \mathbb N| = |A|$, found here.