Cardinality of all infinite subsets with an infinite complement. Working in $\text{ZF}$...
Let $X$ be an infinite set with a given well-ordering relation $\le$.
Define
$\tag 1 \mathcal B(X) = \{ S \in \mathcal P(X) \, | \, S \text{ is infinite } \text{ and } X \setminus S  \text{ is infinite } \}$
I want to show that the cardinality of  $\mathcal B$ is equal to $2^X$. 
If this isn't possible, then can it be demonstrated if we add in the axiom of choice?
My Work
It seems intuitive and 'nice' but I can't find the answer 'searching around' for the 'argument pieces'. I looked over
The cardinality of the set of all finite subsets of an infinite set
but not sure how to proceed. It seems like a 'deja vu' question, but...
 A: Just to make some preliminary comments that I'm sure you're aware of: 
If $X$ and $\mathcal{P}(X)$ are both well-orderable (e.g. if we assume choice), then it's easy, since $\mathcal{P}(X)$ can be partitioned as $\mathcal{P}_{\text{fin}}(X) \sqcup \mathcal{B}(X) \sqcup \mathcal{P}_{\text{cof}}(X)$, where $\mathcal{P}_{\text{fin}}(X)$ and $\mathcal{P}_{\text{cof}}(X)$ are the sets of finite and cofinite subsets of $X$, respectively. So $2^{|X|} = |X| + |\mathcal{B}(X)| + |X| = \max(|\mathcal{B}(X)|, |X|)$, and we must have $|\mathcal{B}(X)| = 2^{|X|}$. 
Without choice, we may have $\mathcal{B}(X) = \emptyset$ even when $X$ is infinite (i.e. $X$ is an amorphous set). 
But in your question, you assume that $X$ is well-orderable, which saves the day. 
We have $\mathcal{B}(X)\subseteq \mathcal{P}(X)$, so in order to show $|\mathcal{B}(X)| = 2^{|X|}$, it suffices by Cantor–Schröder–Bernstein to find an injective function $h:\mathcal{P}(X) \hookrightarrow \mathcal{B}(X)$. Since $X$ is infinite and well-orderable, we can write $X$ as a disjoint union $X = Y\sqcup Z$, where $|X| = |Y| = |Z|$ (think of the "even" and "odd" ordinals). Let $f\colon X\to Y$ and $g\colon X \to Z$ be bijections. Then for any $S\in \mathcal{P}(X)$, let $h(S) = f(S) \cup g(X\setminus S)\in \mathcal{B}(X)$. 
