I am trying to understand this proof of the existence of an uncountable ordinal. I don't see why $\mathcal{P}(\omega \times \omega)$ contains a copy of every countable ordinal as it is said.
For example, what element of $\mathcal{P}(\omega \times \omega)$ would correspond to $\omega\cdot 2$ ?
Let $\alpha$ be a countable ordinal, and let $f\colon\omega\to\alpha$ be a bijection. Then the relation $\{\langle m,n\rangle\mid f(m)\in f(n)\}$ is a well-ordering of $\omega$ with order type $\alpha$ (with $f$ being the isomorphism), and it is an element of $\mathcal P(\omega\times\omega)$.
If you want a particular relation which is isomorphic to $\omega\cdot2$, take the following: $$\{\langle m,n\rangle\mid (m<n\text{ and }m\equiv n\pmod 2)\lor m\text{ is odd, and }n\text{ is even}\}.$$