Show, that under the conditions
(1) $-\infty<a<b<\infty, f\in C([a,b]\times\mathbb{R}), F\in C([a,b]\times [a,b]\times\mathbb{R})$
(2) $\exists~1>c\geq 0~\forall x\in [a,b]~\forall u,v\in\mathbb{R}:~\lvert f(x,u)-f(x,v)\rvert\leq x\lvert u-v\rvert$
(3) $\exists~L\geq 0~\forall x,y\in [a,b]\forall u,v\in\mathbb{R}:~\lvert F(x,y,u)-F(x,y,v)\rvert\leq L\lvert u-v\rvert$
the non-linear integral equation $$ u(x)+\int_a^xF(x,y,u(y))\, dy=f(x,u(x)),~~~x\in [a,b] $$ has exactly one solution in $C([a,b])$.
My proof is inspired by the proof in our lecture for nearly the same theorem: There the right side only depended on $x$.
There is one problem at the end of my proof, where I cannot continue.
Proof
The idea is to use Banach's fixed-point theorem.
Write $$ u(x)=T(u)(x):=-\int_a^x F(x,y,u(y))\, dy+f(x,u(x)),~~~~~x\in [a,b]. $$ Because of condition (1) it is $T\colon C([a,b])\to C([a,b])$.
Because of (2) and (3), it is
$$ \lvert T(u)(x)-T(v)(x)\rvert\leq L\int_a^x\lvert u(y)-v(y)\rvert\, dy+c\lvert u(x)-v(x)\rvert. $$ Then the next idea is to consider the norm $$ \lVert u\rVert_{\lambda}:=\max_{x\in [a,b]}\left\{e^{-\lambda (x-a)}\lvert u(x)\rvert\right\} $$ for a $\lambda >0$ that will be specified later. This norm is equivalent to the max-norm.
Then:
$$ \lvert T(u)(x)-T(v)(x)\rvert\leq L\int_a^x e^{\lambda (y-a)}e^{-\lambda (y-a)}\lvert u(y)-v(y)\rvert\, dy+c\lvert u(x)-v(x)\rvert\\\leq L\int_a^x e^{\lambda (y-a)}\lVert u-v\rVert_{\lambda}\, dy+\lVert u-v\rVert_{\lambda}\\=\frac{L}{\lambda}(e^{\lambda(x-a)}-1)\lVert u-v\rVert_{\lambda}+\lVert u-v\rVert_{\lambda}\\\leq\frac{L}{\lambda} e^{\lambda (x-a)}\lVert u-v\rVert_{\lambda}+\lVert u-v\rVert_{\lambda} $$
Now I have a problem!
In the proof in our lecture the same way gave
$$ \lvert T(u)(x)-T(v)(x)\rvert\leq \frac{L}{\lambda} e^{\lambda(x-a)}\lVert u-v\rVert_{\lambda} $$ so the next step was to divide by $e^{\lambda (x-a)}$, getting then $$ \lVert T(u)-T(v)\rVert_{\lambda}\leq \frac{L}{\lambda}\lVert u-v\rVert_{\lambda}. $$ So choosing $\lambda>L$ gave the desired uniqueness by Banach.
But here I have the additional summand $\lVert u-v\rVert_{\lambda}$ and I do not see how to continue, in order to get an equivalent inequation $\lVert T(u)-T(v)\rVert_{\lambda}\leq \underbrace{\mbox{constant}}_{<1}\cdot \lVert u-v\rVert_{\lambda}$.
For all $x\in[a,b]$ $$\begin{align*} e^{-\lambda (x-a)}|T(u)(x)-T(v)(x)|&\le L\,e^{-\lambda (x-a)}\int_a^x e^{\lambda (y-a)}e^{-\lambda (y-a)}|u(y)-v(y)|\, dy\\ &\qquad\qquad +c\,e^{-\lambda (x-a)}| u(x)-v(x)|\\ &\le L\,e^{-\lambda (x-a)}\,\|u-v\|_{\lambda}\int_a^x e^{\lambda (y-a)}\, dy+c\,\|u-v\|_{\lambda}\\ &=\Bigl(\frac{L\,e^{-\lambda (x-a)}}{\lambda}(e^{\lambda(x-a)}-1)\Bigr)\|u-v\|_{\lambda}+c\,\|u-v\|_{\lambda}\\ &\le\Bigl(\frac{L}{\lambda}+c\Bigr)\,\|u-v\|_{\lambda}. \end{align*}$$ Since $0<c<1$, $\lambda$ can be chosen large enough to have $$ K=\frac{L}{\lambda}+c<1. $$ Then $$ \|Tu-Tv\|_\lambda\le K\,\|u-v\|_{\lambda} $$ with $0\le K<1$.