I have this theorem in my book:
Suppose that $f$ is continuous in the rectangle $$R = \{(t,u) | |t-t_0|\leq T, |u(t)-u_0|\leq L\} $$ and that $$|f(t,u)|\leq M \text{ if } (t,u)\in R$$ Let $\delta = \min(T,L/M).$ If $u(t)$ is any solutions of $\dot{u} = f(t,u)$, $u(t_0) = u_0$ then $$|u(t)-u_0|\leq L \text { when } |t-t_0|\leq \delta.$$ Suppose in addition $f$ is a lipschitz continuous function of $u$ uniformly in $t$. Then the solution of $\dot{u} = f(t,u)$, $u(t_0) = u_0$ is unique in the interval $|t-t_0|\leq \delta$.
The proof for the first part is:
If a solution $u(t)$ stays inside the interval $|u(t)-u_0|\leq L$, then its derivative is bounded by $M$, so the solution cannot escape the interval in less time than $L/M$. Consider $$D = \{0\leq\eta\leq\delta | |u(t)-u_0|\leq L \text{ for all } |t-t_0|\leq\eta\}. $$ Then $0\in D$, and if $\eta\in D$, then $\eta ' \in D$ for all $0\leq \eta '\leq \eta$. Thus $D$ is a nonempty interval. Moreover, $D$ is closed in $[0,\delta]$ because $u(t)$ is a continuous function of $t$. If $\eta\in D$ and $\eta<\delta$, then $f(t,u(t))\leq M$ for $|t-t_0|\leq \eta$, so $$|u(t)-u_0|\leq \bigg |\int_{t_0}^t f(s,u(s))ds\bigg |\leq M\eta < M\delta = L.$$ Since we have a strict inequality, and $u$ is continuous, it follows that there is an $\epsilon>0$ such that $|u(t)-u_0|\leq L$ when $|t-t_0|\leq \eta + \epsilon$. Thus, $D$ is open in $[0,\delta]$, from which we conclude that $D = [0,\delta]$.
My first questions is on the line:
If a solution $u(t)$ stays inside the interval $|u(t)-u_0|\leq L$, then its derivative is bounded by $M$, so the solution cannot escape the interval in less time than $L/M$. Why does the derivative have to be bounded?
The other two questions are on why is $D$ closed? Also why is it open $D$? And why does one want to show that $D = [0,\delta]$?
Any help and comments would be appreciated. Thank you in advance.
As abaktai said, for your first question you have that $\dot u=f$, so the fact that $f$ is bounded makes $\dot u$ bounded.
For your other questions, you want to guarantee that $|t-t_0|<\delta$ implies that $|u(t)-u_0|\leq L$. The trick in the proof is to use the set $D$. In principle $D$ could consist of only $0$, which wouldn't be useful; we need to show it is a bigger interval. If $D=[0,\delta]$, then the statement of the theorem holds.
So the proof first argues that $D$ is an interval. That it is closed, because $$ D=u^{-1}([u_0-L,u_0+l])\cap [t_0-\eta,t_0+\eta] $$ and $u$ is continuous. Next, to show that $D$ is open, the proof shows that if $\eta\in D$, then there exists $\epsilon>0$ with $(\eta-\epsilon,\eta+\epsilon)\subset D$, so $D$ is open.
So, we know that $D$ is an interval $[0,\delta']\subset[0,\delta]$. If we assume that $\delta'<\delta$, then the argument in the proof shows that $\delta'+\epsilon\in D$, an impossibility. So $D=[0,\delta]$.