Restricted Tukey's Theorem/Strong Form

Theorem
Let $X$ be a set.

Let $\AA$ be a non-empty set of subsets of $X$.

Let $'$ be a unary operation on $X$.

Let $\AA$ have finite character.

For all $A \in \AA$ and all $x \in X$, let either:
 * $A \cup \set x \in \AA$

or:
 * $A \cup \set {x'} \in \AA$

Then for each $A \in \AA$ there exists a $C \in \AA$ such that:
 * $A \subseteq C$

and:
 * for all $x \in X$, either $x \in C$ or $x' \in C$.

Proof
Let $A \in \AA$.

Let:
 * $\BB = \set {B: B \subseteq X \text{ and } A \cup B \in \AA}$

It is to be shown that $\BB$ has finite character:

First suppose that $B \in \BB$ and $F$ is a Definition:Finite Subset subset of $B$.

Then since $B \in \BB$, $B \subseteq X$ and $A \cup B \in \AA$.

We wish to show that $F \in \BB$.

Since $F \subseteq B \subseteq X$:
 * $F \subseteq X$

It remains to be shown that:
 * $A \cup F \in \AA$.

$A \cup F \subseteq A \cup B$.

Let $G$ be a finite subset of $A \cup F$.

Then $G$ is a finite subset of $A \cup B$.

Since $A \cup B \in \AA$ and $\AA$ has finite character, $G \in \AA$.

Thus every finite subset of $A \cup F$ is in $\AA$.

Since $\AA$ has finite character, $A \cup F \in \AA$.

Thus $F \in \BB$.

Suppose instead that $B \subseteq X$ and every finite subset of $B$ is an element of $\BB$.

We wish to show that $B \in \BB$.

In order to do this, we must show that $A \cup B \in \AA$.

Let $F$ be a finite subset of $A \cup B$.

$\AA$ has finite character, $F \cap A \in \AA$.

Since $F \cap B$ is a finite subset of $B$, $F \cap B \in \BB$ by assumption.

Then by the definition of $\BB$:
 * $\paren {F \cap B} \cup A \in \AA$

But $F$ is a finite subset of $\paren {F \cap B} \cup A \in \AA$.

Since $\AA$ has finite character, $F \in \AA$.

As this holds for all finite subsets of $A \cup B$ and $\AA$ has finite character:
 * $A \cup B \in \AA$

Let $B \in \BB$ and $x \in X$.

Then:
 * $B \cup A \in \AA$

so either $B \cup A \cup \set x$ or $B \cup A \cup \set {x'}$ is in $\AA$.

But then $B \cup \set x$ or $B \cup \set {x'}$ is in $\BB$ by the definition of $\BB$.

Thus $\BB$ satisfies the premises of the weak form of the Restricted Tukey's Theorem.

Thus there is a $B \in \BB$ such that for all $x \in X$, either $x \in B$ or $x' \in B$.

Let $C = A \cup B$.

Then since $B \subseteq C$, if $x \in X$ then either $x \in C$ or $x' \in C$.

But since $B \in \BB$:
 * $C = A \cup B \in \AA$