= Solution
Let $X=C([0,T];L^2(\mathbb R))$ and define the map suggested by the <variation-of-constants formula>:
$$
(\Phi u)(t)=U(t,0)u_0+\int_0^tU(t,s)f(s,u(s))\,ds.
$$
The <strong continuity> of the <evolution family> and the continuity of $f$ imply that $\Phi$ maps $X$ into itself. Because $U(t,s)$ is unitary and $\|f(t,u)\|_2\leq C$,
$$
\|\Phi u(t)\|_2\leq\|u_0\|_2+Ct.
$$
Equip $X$ with the <exponentially weighted supremum norm>
$$
\|u\|_\alpha=\sup_{0\leq t\leq T}e^{-\alpha t}\|u(t)\|_2.
$$
This equivalent norm makes $X$ a <Banach space>. Using that $f$ is a <globally Lipschitz function> and that the <unitary operator> $U(t,s)$ preserves the norm,
$$
\begin{aligned}
e^{-\alpha t}\|\Phi u(t)-\Phi v(t)\|_2
&\leq L\int_0^te^{-\alpha(t-s)}e^{-\alpha s}\|u(s)-v(s)\|_2\,ds\\
&\leq\frac{L}{\alpha}\|u-v\|_\alpha.
\end{aligned}
$$
Choose $\alpha>L$. Then $\Phi$ is a <contraction mapping>, so the <Banach fixed-point theorem> gives a unique fixed point $u\in X$. This fixed point is exactly the required <mild solution of an abstract Cauchy problem>:
$$
\boxed{u(t)=U(t,0)u_0+\int_0^tU(t,s)f(s,u(s))\,ds}.
$$
The weighted-norm argument works on the whole prescribed finite interval, so no subdivision of $[0,T]$ is needed.
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