Cardinals are Totally Ordered

Corollary to Zermelo's Theorem
Every set of cardinals is totally ordered under $\le$.

Proof
Let $S$ be a set of cardinals.

From Zermelo's theorem, $S$ is well-ordered.

Let $a, b \in S$.

Consider the subset $X = \left\{{a, b}\right\}$ of $S$.

Since $S$ is well-ordered, $\inf \left({X}\right)$ exists and belongs to $X$.

So either $\inf \left({X}\right) = a$ or $\inf \left({X}\right) = b$.

By definition of infimum, we have that $\inf \left({X}\right) \le a$ and $\inf \left({X}\right) \le b$.

It follows that either $a \le b$ or $b \le a$.

So, by definition, $S$ is totally ordered under $\le$.