[Solved]: Is it possible to always construct a hamiltonian path on a tournament graph by sorting?

I assume that you intend $leq$ to be the reflexive and transitive closure of $E$, or directed reachability, that is $leq$ is really the smallest fixpoint of the equivalence you give: $qquad displaystyle a leq b iff exists, v_0 dots v_n. a to v_o to dots to v_n to b$ with $n geq 0$ and $v to u iff (v,u) in E$. The key argument here is that $leq$ is indeed a total order. If it is, you can use it for sorting, and the result is by definition of $leq$ a Hamiltonian path: consider the result of the sort $v_1 dots v_n$. If there were no edge between $v_i$ and $v_{i+1}$, by $V_s = v_i leq v_{i+1}$ there is $v in V$ such that $v_i leq v leq v_{i+1}$. This contradicts that $leq$ is a (total) order and $V_s$ is sorted with respect to $leq$. Unfortunately, $leq$ is not an order relation: it is not antisymmetric. Consider the small tournament based on $K_4$:
[source] Here, $a leq d$ and $d leq a$ (directed cycles are the culprits!) but $a neq d$. So $leq$ is only a (well-)quasi-order and there is no such thing as a sorted node list.