My question is given at the end of the explanation. Let $K\in{}C([a,b]^{2},\mathbb{R})$ and consider the operator $H:C([a,b],\mathbb{R})\to{}C([a,b],\mathbb{R})$ defined by $$H[x](t):=\int_{a}^{t}K(t,\eta)x(\eta)\mathrm{d}\eta.$$ Let $f\in{}C([a,b],\mathbb{R})$ and consider the equation $$x=H[x]+f,\quad\text{on}\ [a,b].$$ Then, $$x=H\big[H[x]+f\big]+f=H^{2}[x]+H[f]+f,$$ which yields by repeating in this manner that $$x=H^{n}[x]+\sum_{k=0}^{n-1}H^{k}[f],{\quad}n=1,2,\cdots\tag{A}$$ with the convention that $H^{0}[f]:=f$. As $K$ is continuous, it is bounded on $[a,b]^{2}$, say $|K(t,s)|\leq{}M$ for all $s,t\in[a,b]$, we can show by induction that $$|H^{n}[x](t)|\leq{}M^{n}\int_{a}^{t}\frac{(t-\eta)^{n-1}}{(n-1)!}|x(\eta)|\mathrm{d}\eta$$ for all $t\in[a,b]$ and $n=1,2,\cdots$. This yields that the operator satisfies $\|H^{n}[x]\|_{\infty}\leq{}M^{n}\frac{(b-a)^{n}}{n!}\|x\|_{\infty}$ for $n=0,1,\cdots$, i.e., $\|H^{n}\|_{\text{op}}\leq{}M^{n}\frac{(b-a)^{n}}{n!}$ for $n=0,1,\cdots$. Therefore, $$\rho(H)=\lim_{n\to\infty}\sqrt[n]{\|H^{n}\|_{\text{op}}}\leq\lim_{n\to\infty}M\frac{(b-a)}{\sqrt[n]{n!}}=0$$ showing that $\rho(H)=0$, which is less than $1$. Then, we can let $n\to\infty$ in (A) and get $$x=\sum_{k=0}^{\infty}H^{k}[f].$$
The point I don't understand here is how $\rho(H)<1$ allows us to let $n\to\infty$ in (A)!
Explanation for user2139: Clearly, for $n=1$, we have $$|H[x](t)|\leq\int_{a}^{t}|K(t,\eta)||x(\eta)|\mathrm{d}\eta\leq{}M\int_{a}^{t}|x(\eta)|\mathrm{d}\eta,$$ which shows that the claim is true. Suppose that the claim is true for some $n$, then we compute that \begin{align} |H^{n+1}[x](t)|={}&\Biggl|\int_{a}^{t}K(t,\eta)H^{n}[x](\eta)\mathrm{d}\eta\Biggr| \leq{}\int_{a}^{t}|K(t,\eta)||H^{n}[x](\eta)|\mathrm{d}\eta\\ \leq{}&M\int_{a}^{t}\Biggl|M^{n-1}\int_{a}^{\eta}\frac{(\eta-\xi)^{n-1}}{(n-1)!}|x(\xi)|\mathrm{d}\xi\Biggr|\mathrm{d}\eta\\ ={}&M^{n}\int_{a}^{t}\int_{a}^{\eta}\frac{(\eta-\xi)^{n-1}}{(n-1)!}|x(\xi)|\mathrm{d}\xi\mathrm{d}\eta\quad\text{(change integration order)}\\ ={}&M^{n}\int_{a}^{t}\int_{\xi}^{t}\frac{(\eta-\xi)^{n-1}}{(n-1)!}|x(\xi)|\mathrm{d}\eta\mathrm{d}\xi\quad\text{(make substitution)}\\ ={}&M^{n}\int_{a}^{t}\frac{(t-\xi)^{n}}{n!}|x(\xi)|\mathrm{d}\xi, \end{align} which shows that the claim is also true for $(n+1)$. By induction, the claim holds for all $n\in\mathbb{N}$.
You have $$ 0=\rho(H)=\lim_{n\to\infty}\|H^n\|^{1/n}. $$ This implies that $\|H^n\|^{1/n}\to0$. So for all $n$ big enough, $\|H^n\|^{1/n}<1/2$, say.
Then your series converges by comparison:
$$ \|\sum_{k=m}^n H^k[f]\|\leq\|f\|\,\sum_{k=m}^n\|H^k\|\leq\|f\|\,\sum_{k=m}^n2^{-k}\leq\frac{\|f\|}{2^{m-1}}, $$ showing that the series is Cauchy.