If $A$ is infinite set then it is partitionable into denumerable (infinitely countable) sets Let $A$ be a non-empty set, such that $\mathbb{N}\preceq A$. Prove that this implies that $A$ is partitionable into denumerable sets.
A hint to this question says that Zorn's lemma is required to prove it. But I'm not very confident when it comes to Zorn's lemma. So I would appreciate your input.
Here's what we are given:
That $\aleph_0 \le |A|$, which means that $A$ is infinite. Also, we are given that there exists an injection from $\mathbb{N}$ to $A$.
Let $\{P_i\}_{i\in I}$ be a partition of $A$. What needs to be shown here is that $P_i$ can be denumerable sets, even if $A$ is uncountable. Now, to apply Zorn's lemma, it is necessary to show that each $P_i$ can be split into chains which have upper bounds. This will lead to the fact that each $P_i$ has a maximal element.
So the questions that I have are the following:
(1) If each $P_i$ has a maximal element, how does this indicate that $P_i$ is denumerable? E.g., $(0,1]$ has the maximal element, $1$, but it is not denumerable.
(2) What is a good example of a chain contained in $P_i$? My guess is that these can be nested sets. For example, one could create a Cantor-like set contained in $P_i$.
Would appreciate your clarification.

I think there's a rather straightforward approach that doesn't need ordering here. We just consider two cases: 1) if $A$ is infinitely countable then we simply break it down into singletons; 2) if $A$ is uncountable, let $B$ be an infinitely countable subset of $A$, then $\{b\},b\in B$ together with $A\setminus B$ is a desired partition. 
