This question is long but the basic idea is super simple: show that $|A| = |B|, |B| = |C| \Rightarrow |A| = |C|$. So please don't be intimidated by the length of this post.
First some definition:
A set $W$ is a connection of two disjoint sets $A$ and $B$ (according to Zermelo) if the following hold:
- $Z \in W \Rightarrow (\exists x \in A, y \in B)[Z = \{x, y\}]$
- For each $x \in A$, there is exactly one $y \in B$ such that $\{x, y \} \in W$
- For each $y \in B$, there is exactly one $x \in A$ such that $\{x, y \} \in W$
For the sake of clarity, we will refer to each element (which is itself a set) as a "pair set" whenever needed; this is done to emphasize that the connection $\Sigma(A,B)$ is not the elements of the form $\{x,y\}$ satisfying condition 2 and 3 above, but the SET of all such elements.
In other words, $\Sigma(A,B)$ refers to an entire set of "pair sets" $\{x,y\}$ satisfying conditions 2 and 3. For example, $\Sigma(\{a,b,c\}, \{a,d\}) = \emptyset$ while $\Sigma(\mathbb N, \{1,2,3,...\})$ could be either the set of all "pair sets" of the form $\{n, n+1\}$ or the set of all "pair sets" $\{0,4\}, \{1, 3\}, \{2, 2\}, \{3,1\}, \{n, n+1\}$ with $n \geq 4$ (as an aside, notice also that $\{1,3\}$ and $\{3,1\}$ are actually the same element). Of course, $\Sigma(\mathbb N, \{1,2,3,...\})$ could be neither of the two. Hence there can be multiple options for $\Sigma(A, B)$ if $A$ and $B$ have same cardinality.
We may assume that for any two disjoint sets $A, B$, there always exists a connection of the two, denoted $\Sigma(A,B)$.
Intuitively, $\Sigma(A,B)$ is nonempty if and only if they have same cardinality.
Now I wish to prove the last part of $x3.3$:
Thanks for all the comments. I believe that the book has a typo and it should read "...and nonempty connections" exist (see also the last sentence of next paragraph).
It's also worth mentioning that $\Sigma(A,B)$ is not a function of $A,B$ (example provided earlier with $\mathbb N$ and $\{1,2,3,...\}$); in other words, we are not sure what exactly it is referring to, although we only know the property of its elements (as an analogy, we know what "a set of $5$ real numbers, each of which is greater than $2816$" is referring to, but there are several types of such a set - it could be $\{10001, 10002, 10003, 10004, 10005\}$ or $\{ \pi^{100}, 19378, 10001, 48888, 2817\}$). Hence, the equivalence relation actually should be interpreted as "$...\exists \Sigma(A,C) \neq \emptyset, \exists \Sigma(B,C) \neq \emptyset$."
The only thing I am having difficulty is the last of the equivalent condition, transitivity. Here it is difficulty since, though the result is intuitively clear, I am not allowed to create bijection between sets and am only allowed to use the existence of class and first six of $ZF$ Axioms, namely axiom of extensionality, emptyset and pairset axioms, separation axiom, powerset axiom, unionset axiom, and axiom of infinity.
Say $D$ and $E$ are appropriate sets "connecting" the pairs $A,B$ and $B,C$, respectively. If $D=E$ then of course we are done, but I am lost on what to do if that is not the case.
The difficulty here is to find some way to construct a set that is "obviously" in bijective correspondence with $A$ and yet disjoint from both $A$ and $C$.
The same difficulty arises if you want to show that $A\sim_{\rm Z} A$.
Without "cheating" and defining Kuratowski pairs so we can just proceed in a more modern way, and also without Regularity, the best I can think of is to diagonalize our way to some new elements. The core of that would be the following lemma -- I'll leave it to you to apply it in your particular situation...
Lemma. Let $X$ be a set. There exists a $Y$ disjoint from $X$ such that $\Sigma(X,Y)$ is nonempty.
Proof. Choose and fix two sets $0$ and $1$ that are not elements of $X$. (That has to be possible, or Separation would create a contradiction through Russell's paradox). Then define $$ \overline X = \{ \{0,x\} \mid x\in X\land \{0,x\}\notin x \} $$ and then set $$ Y = \{ \overline X \cup \{ \{1,x\} \} \mid x\in X \} $$
Even though it looks like we need Replacement, these in fact both exist due to Separation because they are both are subsets of $\mathcal P(X\cup\{0,1\})$.
But no $y\in Y$ can equal any $x\in X$, because $\{0,x\}\in y$ iff $\{0,x\}\in \overline X$ which is exactly if $\{0,x\}\notin x$.