$\DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Res}{Res} \DeclareMathOperator{\VP}{V.P.} \newcommand{\ph}{\varphi} \newcommand{\inth}{\int_{-h}^0} \newcommand{\teta}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\<}{\leq} \renewcommand{\>}{\geq} \newcommand{\Int}{\int\limits} \newcommand{\Sum}{\sum\limits} \newcommand{\Max}{\max\limits} \newcommand{\Sup}{\sup\limits} \newcommand{\Lim}{\lim\limits} \newcommand{\lmin}{\lambda_{\min}} \newcommand{\eqdef}{\stackrel{\mathrm{def}}{=}} \newcommand{\PC}{PC([-h,0],\R^n)} \newcommand{\mh}{\mathfrak{h}} \newcommand{\bh}{\mathbf h} \newcommand{\bN}{\mathbf N} \newcommand{\eps}{\varepsilon} \newcommand{\dt}{\mathrm{d}t} \newcommand{\dd}{\mathrm{d}} \newcommand{\x}{\textup{x}} \newcommand{\y}{\textup{y}} \newcommand{\w}{\textup{w}}$ Consider the delay differential system, $$ \dot{x}(t)=\sum_{k=0}^m \left(A_k x(t-h_k)+\int\limits_{-h_k}^0 G_k(\theta)x(t+\theta)d\theta\right)\quad \forall t\geq t_0,\tag{1} $$ where \begin{gather} t_0\geq 0,~A_k\in\mathbb{R}^{n\times n},~G_k\in C\left([-h_k,0],\mathbb{R}^{n\times n}\right),~k=\overline{0,m},~n,m\in\mathbb{N}, \\ n>0,~m>0=h_0<h_1<\ldots <h_m=h. \end{gather} Let us denote an initial function by $\phi\in \PC$: $$ x(t_0+\theta,\phi)=\phi(\theta)\quad\forall\theta\in[-h, 0],\tag{2} $$ and a system state associated with a moment $t\geq 0$ as $$ x_t(\phi):\theta\mapsto x(t+\theta,\phi)\quad\forall\theta\in[-h, 0]. $$ Consider the following norm in the space $\PC$: $$ \lVert\phi\rVert_h=\sup\limits_{\theta\in [-h,0]}\lVert\phi(\theta)\rVert. $$ Denote the functional $f:\PC\to\mathbb{R}^n$ as $$ f(\phi)=\sum_{k=0}^m \left(A_k \phi(-h_k)+\int\limits_{-h_k}^0 G_k(\theta)\phi(\theta)d\theta\right). $$ Also denote $$ M=\sum_{k=0}^m \left(\left\lVert A_k\right\rVert + \int\limits_{-h_k}^0 \left\lVert G_k(\theta)\right\rVert d\theta\right). $$
I wonder why the following result occurs.
Theorem. For any $t_0\geq 0$ and $\phi\in \PC~\exists~\tau>0$: the system $(1)$ admits a unique solution $x(t,\phi)$ of the initial value problem $(2),$ and the solution is defined on the segment $[t_0-h,t_0+\tau].$
Now I have the following proof attempt.
Proof. It is seen that \begin{align} \lVert f(\phi_1)-f(\phi_2) \rVert =& \Bigg\lVert\sum_{k=0}^m \bigg(A_k \left(\phi_1(-h_k)-\phi_2(-h_k)\right) \\ &+\int_{-h_k}^0 G_k(\theta)\left(\phi_1(\theta)-\phi_2(\theta)\right)d\theta\bigg)\Bigg\rVert\nonumber \\ \leq & \sum_{k=0}^m \left(\left\lVert A_k\right\rVert \lVert\phi_1-\phi_2\rVert_h + \int_{-h_k}^0 \left\lVert G_k(\theta)\right\rVert\lVert\phi_1-\phi_2\rVert_h d\theta\right)\nonumber \\ =&M\lVert\phi_1-\phi_2\rVert_h\quad\forall \phi_1,\phi_2\in\PC.\tag{3} \end{align} Now select $\tau>0$ such that $$ \tau M<1.\tag{4} $$ Define the set \begin{align*} U=\left\{u:[t_0-h,t_0+\tau]\to\mathbb{R}^n,~u(t_0+\theta)=\phi(\theta)\quad\forall\theta\in[-h,0],~u\in C\left([t_0,t_0+\tau],\mathbb{R}^n\right)\right\}, \end{align*} and $\forall u\in U$ define the operator $$ \mathcal{A}(u)(t)= \begin{cases} \phi(t-t_0),\quad t\in[t_0-h,t_0], \\ \phi(0)+\int\limits_{t_0}^t f(u_s)ds,\quad t\in[t_0,t_0+\tau], \end{cases} $$ with $u_s:\theta\to u(s+\theta)~\forall\theta\in [-h,0].$ Having an arbitrary $u\in U,$ let us show that $\mathcal{A}(u)\in U.$ Indeed, \begin{align} \theta\in[-h,0]\implies &t_0+\theta\in[t_0-h,t_0]\implies\mathcal{A}(u)(t_0+\theta)=\phi(t_0+\theta-t_0)=\phi(\theta), \\ u\in PC\left([t_0-h,t_0+\tau],\mathbb{R}^n\right)\implies & g(s)=f(u_s)=\sum_{k=0}^m \left(A_k u(s-h_k)+\int\limits_{-h_k}^0 G_k(\theta)u(s+\theta)d\theta\right)\in PC\left([t_0,t_0+\tau],\mathbb{R}^n\right) \\ \implies &\mathcal{A}(u)(t)=\phi(0)+\int\limits_{t_0}^t f(u_s)ds\in C\left([t_0,t_0+\tau]\right). \end{align} Consequently, $$ u\in U\implies \mathcal{A}(u)\in U.\tag{5} $$ Observe that $(4)$ implies $\forall u^1,u^2\in U$ that \begin{align} &\left\lVert\mathcal{A}(u^1)(t)-\mathcal{A}(u^2)(t)\right\rVert= \begin{cases} 0,\quad t\in[t_0-h,t_0], \\ \begin{aligned} \left\lVert\int\limits_{t_0}^t \left(f(u_s^1)-f(u_s^2)\right)ds\right\rVert\leq &\int\limits_{t_0}^{t_0+\tau} \left\lVert f(u_s^1)-f(u_s^2)\right\rVert ds\leq M\int\limits_{t_0}^{t_0+\tau} \left\lVert u_s^1-u_s^2\right\rVert_h ds \\ \leq &\tau M\sup\limits_{s\in [t_0-h,t_0+\tau]}\left\lVert u^1(s)-u^2(s)\right\rVert,\quad t\in[t_0,t_0+\tau], \end{aligned} \end{cases} \\ &\implies \sup\limits_{s\in [t_0-h,t_0+\tau]}\left\lVert\mathcal{A}(u^1)(t)-\mathcal{A}(u^2)(t)\right\rVert\leq \tau M\sup\limits_{s\in [t_0-h,t_0+\tau]}\left\lVert u^1(s)-u^2(s)\right\rVert.\tag{6} \end{align} Let us show now that $U$ is closed. Assume that $\{u_n \}_{n=1}^{\infty}\subset U,~\exists\lim\limits_{n\to\infty}u_n=u_0.$ So \begin{align} u_n(t_0+\theta)=\phi(\theta)\quad\forall\theta\in[-h,0]\implies u_0(t_0+\theta)=\phi(\theta), \end{align} $C\left([t_0,t_0+\tau],\mathbb{R}^n\right)$ with sup-norm is complete subspace of the complete space of bounded real-valued functions with sup-norm \begin{align} \implies &C\left([t_0,t_0+\tau],\mathbb{R}^n\right)\text{ is closed} \\ \implies & u_0\in C\left([t_0,t_0+\tau],\mathbb{R}^n\right). \end{align} Therefore $U$ is closed. Also $U$ is the subspace of the complete space $C\left([t_0-h,t_0+\tau],\mathbb{R}^n\right)$ with sup-norm. So $U$ is complete. Therefore $(4)$-$(6)$ and Banach Fixed-Point Theorem imply that \begin{align} \exists !~u^*\in U: u^*(t)=\mathcal{A}(u^*)(t)= \begin{cases} \phi(t-t_0),\quad t\in[t_0-h,t_0], \\ \phi(0)+\int\limits_{t_0}^t f(u_s^*)ds,\quad t\in[t_0,t_0+\tau]. \end{cases} \end{align} Denote \begin{align} S=&\left\{s\in[t_0,t_0+\tau]:~s\text{ is discontinuity point of }g(s)=f(u_s)\right\}, \\ s_0(s)=&\left\{\begin{aligned} \Max_{\substack{y\in S \\ y<s}}y,\quad \text{if }\exists\Max_{\substack{y\in S \\ y<s}}y, \\t_0,\quad\text{otherwise}. \end{aligned}\right. \end{align} Assume that $t\in T=\left\{s\in[t_0,t_0+\tau]:~g(s)=\Lim_{y\to s^-}g(y),~g(s_0(s))=\Lim_{y\to s_0(s)^+}g(y)\right\}.$ So $g\in C\left([s_0,t],\mathbb{R}^n\right)$ and Leibniz integral rule gives that \begin{align} \exists !~u^*\in U: u^*(t)=&\phi(t-t_0),\quad t\in[t_0-h,t_0], \\ \dot{u}^*(t)=&\frac{d}{dt}\left(\phi(0)+\int\limits_{t_0}^{s_0} f(u_s^*)ds+\int\limits_{s_0}^{t} f(u_s^*)ds\right)=f(u_t^*),\quad t\in T.\tag{7} \end{align} It is not hard to check that any solution $x$ of $(1)$-$(2)$ belongs $U$: \begin{align} (2)\implies &x(t_0+\theta)=\phi(\theta)\quad\forall\theta\in[-h,0], \\ (1)\implies &\exists\dot{x}\quad\forall t>t_0,~\exists\dot{x}_+(t_0) \\ \implies &x\in C\left([t_0,t_0+\tau],\mathbb{R}^n\right). \end{align} Therefore $x\in U$ and the proof is complete if $g(s)=f(u_s)\in C\left([t_0,t_0+\tau],\mathbb{R}^n\right).$
Notice that if $(7)$ is fulfilled $\forall t\in [t_0,t_0+\tau]\setminus T$ then the proof is totally complete. But I don't know how to show it. Maybe there is another approach to the problem?