I come up with this theorem when I prove: If $A$ is uncountable and $B$ in non-empty, then $A \times B$ is uncountable. I found that proving this is not so hard by creating a bijection $f:A \times \{b\} \to A$ where $b \in B$ and notice that $A \times \{b\} \subseteq A \times B$. But I'm not sure if my proof for the generalization contains any error.
Please help me check the part that I define mapping $G$!
Lemma: If $f:A \to B$ is surjective and $A$ is countable, then $B$ is countable.
The Cartesian product of a family $(A_i\mid i\in I)$ is defined as $$\prod\limits_{i\in I}A_i=\{f:I\to\bigcup A_i\mid f(i)\in A_i \text{ for all } i \in I\}$$
Theorem: Let $(A_i \mid i \in I)$ be the family of non-empty indexed sets where $A_k$ is uncountable for some $k \in I$. Then $\prod\limits_{i\in I}A_i$ is uncountable.
We define a mapping $G:\prod\limits_{i\in I}A_i \to A_k$ by $G(f)=f(k) \in A_k$ for all $f \in \prod\limits_{i\in I}A_i$. Then $G$ is clearly surjective. Assume the contrary that $\prod\limits_{i\in I}A_i$ is countable. Then, by Lemma, $A_k$ is countable, which is a contradiction. Hence $\prod\limits_{i\in I}A_i$ is uncountable.
Update: On the basis of Kavi Rama Murthy's and Saucy O'Path's comments, I added the proof that $G$ is surjective here.
By Axiom of Choice, there exists a mapping $f:I\to\bigcup A_i$ such that $f(i)\in A_i$ for all $i \in I$. For any $a \in A_k$, we define a mapping $f_a$ by $f_a(i)=f(i)$ for all $i \neq k$ and $f_a(k)=a$. Then $f_a$ is clearly in $\prod\limits_{i\in I}A_i$ and $G(f_a)=a$. Hence $G$ is surjective.
Best Answer
Claiming that $G$ is surjective is a slightly delicate matter, because it requires you to prove that for all $u\in A_k$ there is a function $f:I\to \bigcup_{i\in I} A_i$ such that $f(i)\in A_i$ for all $i\in I$ and $f(k)=u$. This is certainly false if $A_i=\emptyset$ for some $i\ne k$ (and the entire thing is false in that instance). Giving a proper argument for the case where $A_i\ne \emptyset$ for all $i$ is tantamount to the axiom of choice.