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.