Prove that $|Z(A_i)|=|Z(B_i)|$. We denote $ | A | $ as the cardinality of the set $ A $.

Let $\{A_i\}_{i\in i}$ be an indexed family of classes; let $$A=\bigcup_{i\in I}A_i$$ the product of the classes $A_i$ is definded to be the class $$Z(A_i)=\{f: f:I\to A\, \text{is a function, and}\, f(i)\in A_i, \forall i\in I\}$$

Let $\{A_i\}_{i\in I}$ and $\{C_i\}_{i\in I}$ be families of sets. If $|A_i|=|C_i|$ for each $i\in I$, prove that $$\left|Z(A_i) \right |=\left |Z(C_i) \right |.$$
If we make $ A =\displaystyle\bigcup_{i\in I}A_i $ and $ C = \displaystyle\bigcup_{i\in I}C_i$, we can have $ | A | = | C | $. Besides that there is a bijection $ f_i: A_i \to C_i $. But I don't know how to define the bijection from $ Z(A_i) $ to $ Z(C_i)$.
 A: Since $|A_i|=|C_i|$, the set $H_i=\{h:A_i\to C_i\mid h\text{ is a bijection}\}$ is nonempty for each $i\in I$. The Axiom of Choice states that the product $Z(H_i)$ is nonempty, thus we can pick some $h\in Z(H_i)$. Let's use the notation $h_i=h(i)$, then each $h_i:A_i\to C_i$ is a bijection.
Now, for any $f\in Z(A_i)$, let $g_f:I\to \bigcup C_i$ be the function defined as $i\mapsto h_i(f(i))$, which is a function in $Z(C_i)$: if $i\in I$, then $f(i)\in A_i$, thus $h_i(f(i))\in C_i$.
Now the map $f\mapsto g_f$ is a bijection $Z(A_i)\to Z(C_i)$. I'll leave it to you to check injectivity and surjectivity.

As for the use of the Axiom of Choice, this is indeed necessary. A famous example is the "pair of socks" versus "pair of shoes", due to Bertrand Russell: it is possible to choose one shoes from an infinite number of pairs of shoes (one simply takes the left shoe each time), but it's not possible to choose one sock from an infinite number of pairs of socks (since "left" socks and "right" socks are indistinguishable).
More mathematically speaking, one can take the set of indices $I=\Bbb N$ to be the natural numbers, and let $A_i=\{2i,2i+1\}$ for each $i\in \Bbb N$. Since we can always just pick the smaller of the two numbers, we know that $Z(A_i)$ is nonempty (in fact $|Z(A_i)|=2^{|\Bbb N|}=2^{\aleph_0}$).
Meanwhile, it is possible to define sets $C_i=\{x_i,y_i\}$ where $x_i\neq y_i$, such that it is impossible to formulate a criterium that distinguishes $x_i$ from $y_i$ for all $i$ simultaneously. For example, I can take $x_i,y_i$ to be two distinct sets of real numbers for each $i\in \Bbb N$. In the absence of the Axiom of Choice, it is consistent that there are no choice functions on the pairs $C_i$, hence $Z(C_i)=\varnothing$.
