Let $K\in C[a,b]^2, y \in C[a,b]$ and $\lambda \in \Bbb R$ with $\lambda \neq 0$. Show that the integral equation $$x(t) = \lambda \int_a^b K(t,\tau)x(\tau)d\tau +y(t), t\in[a,b]$$ has a unique solution in C[a,b] for sufficiently small $\lambda$.
For the solution I think I should prove that the operator $T:x \to \lambda \int_a^b K(.,\tau)x(\tau)d\tau +y(.)$ is a contraction. But I do not know how to prove it.
On
In order to be able to use the Banach fixed-point theorem we need a complete metric space, so I assume that $C([a,b])$ is endowed e.g. with the norm $\| \cdot \|_\infty$.
I assume wlog $\lambda \ge 0$; thus we have $$ \begin{split} |T(x_1(t)) - T(x_2(t))| & \le \lambda \int_a^b |K(t,\tau) (x_1(t) - x_2(t))| \, d \tau \\ & \le \lambda (b-a) \|K\|_{L^\infty ([a,b] \times [a,b])} \| x_1 - x_2 \|_{L^\infty ([a,b])}. \end{split} $$Picking $\lambda < 1/(\|K\|_{L^\infty ([a,b] \times [a,b])} |b-a|)$ will give you a contraction, and therefore the existence of the unique fixed point $Tx=x \in C([a,b])$.
You first have to equip the function space $C[a,b]$ with the right topology. We'll pick the obvious one which is the uniform convergence topology, with norm $$ \|f\|_\infty := \sum_{x\in[a,b]}|f(t)| $$ Which makes $C[a,b]$ a Banach space. We denote by $\|K\|_\infty$ in the same way the upper bound of $K$'s values over $[a,b]^2$.
That operator $T$ is a map $C[a,b]\to C[a,b]$ is obvious from the dominated convergence theorem and the fact we are integrating other a compact interval.
For any $x_1$ and $x_2$ in $C[a,b]$ and $t\in[a,b]$, $$ |(Tx_2-Tx_1)(t)| = |\lambda| \left|\int_a^b K(t,\tau)(x_2(\tau)-x_1(\tau))\,d\tau\right| \leq |\lambda|\int_a^b |K(t,\tau)||x_2(\tau)-x_1(\tau)|\,d\tau \leq |\lambda|(b-a)\|K\|_\infty\|x_2-x_1\|_\infty $$ Thus, for $|\lambda|$ strictly smaller than $\frac{1}{(b-a)\|K\|_\infty}$, $T$ is a contraction on the entire Banach space $C[a,b]$, and the Banach fixed-point theorem yields the existence and unicity of a function $x\in C[a,b]$ such that $$ x = Tx. $$