Solution
= Solution
Choose $A>0$ and set
$$
M=\max_{|x|\leq A}|\phi(x)|.
$$
Take
$$
T_1\leq\frac A{\max\{M,1\}}.
$$
Inductively, if $|f_n(s)|\leq A$ on $[0,T_1]$, then
$$
|f_{n+1}(t)|
\leq\int_0^t|\phi(f_n(s))|\,ds
\leq MT_1\leq A.
$$
Since $f_0=0$, every iterate is bounded by $A$. Moreover,
$$
|f_{n+1}(t)-f_{n+1}(s)|
=\left|\int_s^t\phi(f_n(r))\,dr\right|
\leq M|t-s|.
$$
\b[Thus $(f_n)_{n\geq1}$ is uniformly bounded and uniformly Lipschitz, hence equicontinuous, on $[0,T_1]$.]