I want to prove that if $A$ is a D-Infinite set (i.e. it contains a countable subset $X$), then the set of the infinite parts of $A$, $P_{\infty}(A)$ has the same cardinality of $P(A)$.
I know that if I have sets $|B|=|C|$ such that $|B \cup C|=|B|$; $B \subset D$ and $C \subset (D-B)$ then $|D|=|D-B|$.
How can I apply this to the initial theorem?
I can consider two cases:
Case 1 = $A$ is countable. Then the proof is simple.
Case 2 = $A$ is more than countable. And here I'm stuck!
Can I do this without the Axiom of choice or the Countable Axiom of choice? Thank you.
You don't need the axiom of choice for this. You just need the Cantor-Bernstein theorem, which is provable without using choice at all.
Let $X\subseteq A$ be a countably infinite set such that $|A\setminus X|=|A|$, such $X$ exists since we can pick any countable subset of $A$, index it using $\Bbb N$, and take $X$ those with even indices.
Since $|A|=|A\setminus X|$ we have that $|\mathcal P(A)|=|\mathcal P(A\setminus X)|$. If $B\subseteq A\setminus X$ then $B\cup X$ is an infinite subset of $A$, moreover $B\mapsto B\cup X$ is injective (I leave it for you to check).
So we have an injection from $\mathcal P(A)$ into $\mathcal P_\infty(A)$, and now using the Cantor-Bernstein theorem we're done.