For case 1, I proved that if $A$ is finite/countable, then $|A\times A|$ is finite/countable. However, I don’t know how to proceed with showing $|A\times A|\geq|A|$. Specifically, I want to say for every $a_i$ in $A$, there is a corresponding $(a_i, a_k)$ in $|A\times A|$ whose list is certainly greater than a single $a_i$ for a corresponding $i$. However, I was given a remark which states if $|A| \geq |B|$, then there exists some $f:A \to B$ that is onto. How would I show $f:|A \times A|\to|A|$ is onto?
Case 2 would be for $A$ is not countable.
To show that $f$ is onto, consider a sort of projection. For example, if
$$A = \{1,2\}$$
then
$$A \times A = \{(1,1)\;,\;(1,2)\;,\;(2,1)\;,\;(2,2)\}$$
We can define the function $f : A\times A \rightarrow A $ by
$$f(x,y)=x$$
Then $(1,1)$ and $(1,2)$ map to $1$, and $(2,1)$ and $(2,2)$ map to $2$. Thus, every element of $A$ has a preimage.
Can you see how this might prove $f$ is surjective in a more general case if $A$ is nonempty? (Or "onto," whichever terminology you prefer.) About the only nuance you would really need to consider for this would be $A = \emptyset$, which you can handle separately and more trivially.