Given a set $S$, I want to find a total antisymmetric relation $R$ on $S$:
$\forall i,j\in S$ either $iRj$ or $jRi$ or both.
In the case both, we have $i=j$, therefore any two distinct elements of $S$ have a fixed direction.
The Axiom of Choice/Well-Ordering Principle can work it out. But this is so muck weaker than AC/WO. So I am wondering if this is gonna work in ZF-system.
This would imply a statement which is probably close in strength to the full Axiom of Choice. That would be what we think of informally as the "choice of one sock from each pair in an infinite drawer": given a collection of sets $\{ A_\lambda : \lambda \in \Lambda \}$ such that $|A_\lambda| = 2$ for each $\lambda$, then by the assumption that your statement is true, find a total antisymmetric relation on $\bigcup_{\lambda \in \Lambda} A_\lambda$. Then, for each $\lambda$, we can choose the element of $A_\lambda$ which is smaller than the other in this order, and thus construct a choice function for $(A_\lambda)_{\lambda \in \Lambda}$.
Actually, a very similar argument will work in the more general case that each $A_\lambda$ is a finite set, if you strengthen the statement to assert the existence of a total order on every set (i.e. a partial order which also satisfies the given condition of being total antisymmetric).