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 $X$
Now my question is:
Is the axiom $``\forall X \, \exists C: C \text{ chosen from } X"$ equivalent to the $\sf AC$ over $ \sf Z \text {-Power} $?
I was thinking of defining a function $f$ that sends each element $x$ of $X$ to the set of all maximal overlaps (\non-overlaps) between $x$ and other elements of $X$, that is:
$\begin{align} x^{\cap} =& \{ y \subsetneq x: \exists K \subseteq X ( y= \bigcap K) \land \\&\neg \exists L \exists z( L \subseteq X (z = \bigcap L) \land z \subsetneq y) \}\end{align}$
$ x^\varnothing = \{l : \forall y \in X (l \in y \to y=x)\}$
Where: $\bigcap K = \{y: K \neq \emptyset \land \forall x \in K (y \in x)\}$
So the desired set is $f(x)= x^\cap \cup \{x^\varnothing\}$
If we had replacement then we get the set $f``X=\{f(x): x \in X\}$, take its union then we have a set of disjoint sets, then apply $\sf AC$ on it, and we get the set $C$ chosen from $X$ that meets the account here.
But it appears to me that this argument essentially needs Power set axiom, since if we had that then we can have the required replacement within some iterative power of $X$ and the argument would go through!
The point is to build $x^\cap$ without resorting to Power, and also to be able to have the needed replacements without Power.
However, the above is my personal attack against this problem, which might not be correct.
Your statement is not even a theorem of $\sf ZFC$, so it is certainly not equivalent to $\sf AC$ over $\sf ZF$, let alone $\sf Z-Pwr$. Consider for a rational number $q$, $L_q=\{p\in\Bbb Q\mid p<q\}$, and consider the family of sets $\{L_q\mid q\in\Bbb Q\}$.
Given any $C$ which meets all of them, then $C$ has to be infinite, since it must be unbounded from above, since if $p$ is any rational number, $C$ must contain some $q>p$ in order to meet $L_p$. But now if $p<q$ are both elements of $C$, then $C\setminus\{p\}$ also meets all of the points.
Note that we didn't use any power set of replacement. Indeed, your axiom is simply inconsistent with $\sf Z-Pwr$.