We call any map $f$ a choice function of the family of sets $\mathcal{K}$ iff $f$ is a map from $\mathcal{K}$ to $(\cup_{S\in\mathcal{K}}S)$ such that $\forall X\in\mathcal{K}(f(X)\in X)$ likewise we call $f$ an anti-choice function of $\mathcal{K}$ iff $f$ is a choice function of $\mathcal{K}$ such that $\small\forall X,Y\in\mathcal{K}(X\subset Y\rightarrow f(X)\neq f(Y))\iff \forall X\in\mathcal{K}(A,B\in f^{-1}[\{f(X)\}]\rightarrow A\not\subset B)$.
Now if $\forall X,Y\in\mathcal{K}(X\not\subset Y)$ then all choice functions of $\mathcal{K}$ are 'anti-choice' functions. But what about other families? Does AC imply all families of non-empty sets have anti-choice functions?
The concept is entirely inconsistent.
Take $\{\{0\},\{1\},\{0,1\}\}$ to be a family of non-empty sets. Then any choice function must map $\{0\}$ to $0$ and $\{1\}$ to $1$. But now $\{0,1\}$ cannot be mapped to either of its elements.
For an example with infinite sets, consider all the initial segments of $\Bbb Q$ except $\varnothing$. There are uncountably many of them (each one defining a unique $r\in\Bbb R$, and $\Bbb Q$ itself), so by cardinality reasons alone uncountably many must have chosen the same element. And since the collection as a whole is a chain, "anti-choice" is necessarily injective, which is impossible.