A non-derivative proof.
What we have is $\forall u,v. (u>v\implies f(u)>f(v))$.
We want to prove $\forall u,v. (f(u)>f(v)\implies u>v)$.
Suppose the opposite, that there exist $u_0,v_0$ such that $f(u_0)>f(v_0)$ is true and $u_0>v_0$ is false for the sake of contradiction.
Since $u_0>v_0$ is false, $u_0=v_0$ or $u_0<v_0$.
Case 1. $u_0=v_0$.
Then $f(u_0)=f(v_0)$, contradiction.
Case 2. $u_0 < v_0$. Then since $\forall v,u. (v>u\implies f(v)>f(u))$, we have $f(v_0)>f(u_0)$, contradiction.
Hence there cannot exist such $u_0,v_0$.