The continuum hypothesis (CH) and its equivalent For a set $A \subseteq \mathbb{R}^2$ and $x,y \in \mathbb{R}$, we define $A^y=\{x \in \mathbb{R}\mid (x,y) \in A\}$ and $A_x=\{y \in \mathbb{R}\mid(x,y) \in A\}$.
Proposition: The continuum hypothesis (CH) is equivalent to the existence of a set $A \subseteq \mathbb{R}^2$ such that $A^y$ and $(\mathbb{R} \setminus A)_x$ are both countable for all $x,y \in \mathbb{R}$.
I got stuck in the converse part which is following:
Let $A \subseteq \mathbb{R}$ be as defined in the proposition and suppose that CH fails. Then $\aleph_1 < \mathfrak{c}$. We well order $\mathbb{R}$ as $\{x_{\alpha}\mid\alpha < \mathfrak{c}\}$ and let $X=\cup_{\alpha < \aleph_1} A^{x_{\alpha}}$. By assumption each $A^{x_\alpha}$ is countable, so $\operatorname{card}(X) \leq \aleph_1 < \mathfrak{c}$. It follows that we can find some $x \in \mathbb{R} \setminus X$. Then for every $\alpha < \aleph_1$, $x \notin A^{x_\alpha}$. So $(x,x_\alpha) \notin A$. Hence $x_\alpha \in (\mathbb{R}\setminus X)_x$. Thus $\operatorname{card}((\mathbb{R}\setminus X)_x) \geq \aleph_1$, a contradiction.
I do not understand
(i) Why well ordering $\mathbb{R}$ during the proof?
(ii) Why $\operatorname{card}(X) \leq \aleph_1 < \mathfrak{c}$?   
 A: First of all, note that we need the axiom of choice here.
In the failure of the axiom of choice it is consistent that the real numbers are a countable union of countable sets, but the continuum hypothesis is false (in the sense that there is an uncountable set of real numbers which is not of the same cardinality of the real numbers themselves). In that case write $\Bbb R$ as the disjoint union of $A_n$ for $n\in\omega$, all are countable. And let $A\subseteq\Bbb R^2$ be defined as $$A=\{(x,y)\mid x\in A_n\land y\in A_m\rightarrow n\leq m\}$$
Then for every $y\in\Bbb R$, if $y\in A_n$ then $A^y=\bigcup_{k\leq n}A_k$ which is a finite union of countable sets; and for each $y\in\Bbb R$, we have $y\in(\Bbb R^2\setminus A)_x$ if and only if $(x,y)\notin A$ namely $y\in A_k$ for some $k<n$, so again we have a finite union of countable sets.

So now that we have established that the axiom of choice is essential here, to some degree anyway, it becomes somewhat obvious why we need to well-order $\Bbb R$. This is even more apparent if you consider the implication $\sf CH\implies (*)$, where $(*)$ is the existence of such $A$:
Assume that $\sf CH$ holds, let $\{x_\alpha\mid\alpha<\omega_1\}$ be an enumeration of $\Bbb R$ and let $A=\{(x_\alpha,x_\beta)\mid \alpha<\beta\}$. It is not hard to check that this $A$ indeed satisfies the wanted property.
So the same idea should work in the other direction. But then we have a bit of a problem, since here we only had to show one witness, and in the second direction we need to show that if $\sf CH$ fails, then there are no witnesses.
To solve that, we well-order $\Bbb R$ and show that no such $A$ can exist. And indeed if $A$ is any set with the wanted property then $X=\bigcup_{\alpha<\omega_1}A^{x_\alpha}$ is a union of $\aleph_1$ countable sets, therefore its cardinality is $\aleph_1$.
