Is this Theorem equivalent to Axiom of Choice? From your answer, I know that they are equivalent and I tried to give a proof that AC $\implies$ this Theorem and that this Theorem $\implies$ AC.
Can you please check the correctness of my proofs?
Thank you for your answers so much!
Axiom of Choice:

Let $\mathcal{F}$ is a collection of non-empty disjoint sets. Then there exists a choice function $f:\mathcal{F} \to \bigcup \mathcal{F}$ such that $f(A) \in A$ for all $A \in \mathcal{F}$.

The Theorem:

For any set $\mathcal{F}$, there exists a function $g:\bigcup \mathcal{F} \to \mathcal{F}$ such that $a \in g(a)$ for all $a \in \bigcup \mathcal{F}$.

Proof that AC implies this Theorem:

Let $R(a)=\{A \in \mathcal{F}\mid a \in A\}$ and $W=\{\{a\} \times R(a)\mid a \in \bigcup \mathcal{F}\}$.
Then $W$ is a collection of non-empty disjoint sets. Apply AC, we have a function $f:W \to \bigcup W$ such that $f(\{a\} \times R(a)) \in \{a\} \times R(a) \implies f(\{a\} \times R(a))=(a, A_a)$ where $a \in A_a \in R(a)$.
So the set $\{(a,A_a) \mid a \in \bigcup \mathcal{F}\}$ or equivalently
  $\{f(\{a\} \times R(a)) \mid a \in \bigcup \mathcal{F}\}$ represents the desired function $g$.

Proof that this Theorem implies Axiom of Choice - Approach 1:

Let $W=\{\{\mathcal{F} \times A,a\} \mid a \in A \in \mathcal{F}\}$.
  
  
*
  
*$\mathcal{F} \times A \neq b$ for all $A \in \mathcal{F}$ and $b\in \bigcup \mathcal{F}$.
  
  
  Assume that $\mathcal{F} \times A = b$ where $b \in B \in \mathcal{F}$, and that $a \in A$. Then $b \in B \in \{B\} \in \{\{B\},\{B,a\}\}=(B,a)\in \mathcal{F} \times A=b$. This contradicts to Axiom of Regularity. 

  
*We generate function $f$ as follows:
  
  
  Apply this Theorem, there exists a function $g:\bigcup W \to W$ such that $a \in g(a) \in W$.
$f:\mathcal{F} \to \bigcup \mathcal{F}$ such that $f(A)=g(\mathcal{F} \times A) \setminus \{\mathcal{F} \times A\}$ for all $A \in \mathcal{F}$.

Proof that this Theorem implies Axiom of Choice - Approach 2:

Let $R(x)=\{A \in \mathcal{F} \mid x \in A\} \cup \{\mathcal{F} \times \{x\}\}$ and $W=\{R(x) \mid\ x \in \bigcup \mathcal{F}\}$.
  
  
*
  
*$\mathcal{F} \times \{x\} \neq A$ for all $A \in \mathcal{F}$ and $x \in \bigcup \mathcal{F}$.
  
  
  Assume that $\mathcal{F} \times \{x\} = A$ where $A \in \mathcal{F}$ and $x \in \bigcup \mathcal{F}$. Then $A \in \{A\} \in (A,x) \in \mathcal{F} \times \{x\}=A$. This contradicts to Axiom of Regularity.
Thus $\mathcal{F} \times \{x\}$ is an distinct indicator of $R(x)$. As a result, $R(a) \neq R(b)$ whenever $a \neq b$, or equivalently $R(x)$ is injective for all $x \in \bigcup \mathcal{F}$.

  
*We generate function $f$ as follows:
  
  
  Apply this Theorem, there exists a function $g:\bigcup W \to W$ such that $x \in g(x) \in W$. It is clear that $A \in W$ for all $A \in \mathcal{F}$.
Let $f(A)=R^{-1}(g(A))$ for all $A \in \mathcal{F}$. 

 A: Well. Yes. Because the axiom of choice (for general families) is equivalent to "choice from families of pairwise disjoint sets".
The proof seems okay. 
A: Yes, the principle you've written is equivalent to AC. 
Suppose I have a family $\mathcal{F}$ of nonempty disjoint sets for which I want to construct a choice function. Fix some $x$ such that $\langle x, a\rangle\not\in\bigcup\mathcal{F}$ for any $a\in\bigcup\mathcal{F}$.  For $a\in\bigcup\mathcal{F}$, let $S_a=\{f\in\mathcal{F}: a\in f\}\cup\{\langle x, a\rangle\}$. Note that the second term here means $S_a\not=S_b$ whenever $a\not=b$.
Now consider the set $X=\bigcup_{a\in \bigcup\mathcal{F}}S_a$, and apply your principle to it to get a map $g$ sending each element of $X$ to some $S_a$ containing it. We now argue:


*

*Each $f\in\mathcal{F}$ is in $X$.

*For each $f\in\mathcal{F}$ there is exactly one $a$ such that $g(f)=S_a$. This is just the distinctness of the $S_a$s, which we brought about artificially above.
Let $h(f)$ denote that unique $a$. Then $h$ is a choice function for $\mathcal{F}$.
