Consider the $1$-d ODE $$-u_{xx}+u-\epsilon u^{2}=f, \tag{1}$$ where $f$ is a nice RHS, say $f\in\mathcal{S}(\mathbb{R})$, and $\epsilon>0$. By using the Bessel potential, one looks for solutions to this equation by solving the integral equation $$u(x)=\int_{\mathbb{R}}\frac{1}{2}e^{-|x-y|}[f(y)+\epsilon u^{2}(y)]dy=T(u)(x)$$ It's clear that $T$ maps the space of continuous, bounded functions $X:=C_{b}(\mathbb{R})$ into itself. For $R>0$ fixed and for $0<\epsilon=\epsilon(R)$ sufficiently small, $T: B_{R}\rightarrow B_{R}$ is a contraction since \begin{align*} |u(x)-v(x)|\leq\epsilon\int_{\mathbb{R}}\frac{1}{2}e^{-|x-y|}|u(y)-v(y)|(|u(y)|+v(y)|)dy\leq\frac{\epsilon}{2}(\|u\|_{X}+\|v\|_{X})\|u-v\|_{X}\|e^{-|\cdot|}\|_{L^{1}} \end{align*} So by Banach's fixed point theorem, $T$ has a unique fixed point $u\in X$.
We really want a classical solution to the ODE (1). Since $\frac{1}{2}e^{-|x|}$ is a fundamental solution for $(I-\Delta)$, we see that $$u_{xx}\stackrel{\mathcal{D}'}{=} u-\epsilon u^{2}-f$$ which shows that $u_{xx}$ is continuous. By the equivalence theorem for distributional and classical derivatives, we see that $u_{x}$ is $C^{1}$ and therefore $u$ is $C^{2}$. By iterating this argument, we actually get $u\in C^{\infty}$ and therefore is a classical solution to (1).
I've seen it written where people do the fixed argument in a domain $Y\subset X$, say $Y:=C_{b}^{2}$. It's clear that $T$ will be a contraction on a ball $B_{R}$ for $\epsilon=\epsilon(R)>0$ sufficiently small, but the range of $\epsilon>0$ here will a priori be smaller than in the case $T: X\rightarrow X$. Based on the argument I've given above, I don't see why one would work with this smaller space since it's more restrictive condition on $\epsilon$.