Theorem 4.1. Let $U(t,u) $ be a continuous real valued function on an open $(t,u)$-set here $t,u\in\mathbb{R}$. Let $u=u^0(t)$ the maximal solution of $$u'(t)=U(t,u),\ \ u(t_0)=u_0$$ Let $v(t)$ be continuous function on [t_0,t_0+a] satisfying the condition $v(t_0)\leq u_0,(t,v(t))\in E,$ and $v(t)$ has a right derivative $D_R v(t)$ on $t_0\leq t\leq t_0+a$ such that $$D_Rv(t)\leq U(t,v(t))$$ Then, on a common interval of existence of $u^0(t)$ and $v(t)$, $$v(t)\leq u^0(t)$$
Proof of theorem: It is sufficient to show that there exists a $\delta >0$ such that $v(t)\leq u^0(t)$ holds for $[t_0,t_0+\delta]$. For if this is the case and $u^0(t)$,$v(t)$ are defined on $[t_0,t_0+\beta]$, it follows that the set of $t$-values where $v(t)\leq u^0(t)$ holds cannot have an upper bound different from $\beta$.
From now on, this is my idea, actually I want to clarify above sufficient condition.
So I let $A=\{\theta\in [0,\beta]:v(t)\leq u^0(t) on [0,\theta]\}$
If $\beta\leq \delta $ clearly $\beta \in A$ therefore $\beta$ is unique upper bound of $A$.
Now I consider the case $\beta>\delta$
Let $\eta = \sup A $ for contradiction assume $\eta \in [\delta,\beta)$
Now we can easily show that $v(t)\leq u^0(t)$ on $[t_0,\eta)$ from the definition of supremum. And next actually $\eta \in A$ because if we assume $u^0(\eta)< v(\eta)$ because of continuity of $v$ and $u_0$ there exist $\epsilon >0$ s.t $u^0(t)< v(t)$ hold on $(\eta-\epsilon,\eta+\epsilon)$. This is contradiction.
I'm stuck on here I don't know how to use $D_Rv(t)\leq U(t,v(t))$ to make contradiction.
Please give me some help.