This is actually a fix of the axiom presented to MathStackExchange in a prior posting.
Lets define a binary relation predicate "chosen from" as:
$$\begin{align}C \text{ chosen from } X \iff& \forall x \in X \, (x \cap C \neq \emptyset) \, \land \\&\neg \exists B \subsetneq C : \forall x \in X \, (x \cap B \neq \emptyset)\end{align}$$
In English: Set $C$ is chosen from set $X$ if and only if $C$ has nonempty intersection with every element of $X$ and such that no proper subset of $C$ can have nonempty intersection with every element of $X$
Define
$\begin{align} \text {Selectable} (X) \equiv_{df} & \forall x \in X \ \exists y \subseteq x \\& \big{(}y \neq \emptyset \land \exists K \subseteq X \, ( y= \bigcap K) \land \\&\neg \exists L \exists z \, (z \neq \emptyset \land L \subseteq X \land z = \bigcap L \land z \subsetneq y) \big{)} \end{align} $
Where: $\bigcap K = \{m: K \neq \emptyset \land \forall k \in K (m \in k)\}$
In English: $X$ is Selectable if and only if of every element $x$ of $X$ there is at least a subset of $x$ that is a maximal overlap in $X$ [that is: $x$ is a nonempty intersection of some elements of $X$ but not a proper superset of a nonempty intersection of elements of $X$]. Selectable intuitively means "can be selected from its elements".
Now we stipulate the following axiom:
$\forall X \, (\text{Selectable} (X) \to \exists C: C \text{ chosen from } X) $
Now the questions are:
Is this axiom equivalent to $\sf AC$ over the rest of axioms of Zermelo set theory?
If yes, then is Power set axiom required to prove the equivalence of this axiom with $\sf AC$ over the rest of axioms of Zermelo set theory?