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...

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 $$h(S) = f(S) \cup g(X\setminus S)\in \mathcal{B}(X)$$.