Banach fixed point theorem application

I'm trying to use the Banach fixed point theorem to show that an intergral equation has a unique solution, but can't seem to make my answer work any help would be appreciated.

Let $$f:[a,b] \rightarrow \mathbb{R}$$, $$g:[a,b] \rightarrow \mathbb{R}$$, $$k:[a,b] \times [a,b]$$ be continuous and $$\sup_{x \in[a,b]} \int_{a}^{b} |k(x,y)|$$ = $$p<1$$. Show there exists a unique continuous $$f$$ that solves:

$$f(x)=g(x)+\int_{a}^{b}k(x,y)f(y)$$

My Attempt

Define $$T(f(x))=g(x)+\int_{a}^{b}k(x,y)f(y)$$.

T is a map that takes continuous maps of the form of f to functions of the form of f, with the sup norm this is a banach space so if i prove a contraction I'm done.

$$\|T(f(x))-T(s(x))\| = \|\int_{a}^{b}k(x,y)(f(y)-s(y))\|\leq p\bigg|\int_{a}^{b}(f(y)-s(y))\bigg|$$

By the MVT $$\exists c \in [a,b]$$ s.t.

$$p\bigg|\int_{a}^{b}(f(y)-s(y))\bigg|=p\bigg|(a-b)(f(c)-s(c))\bigg|$$

I want to conclude that this is $$\leq p \|f-s\|$$ or something similar but I don't have that $$(a-b)<1$$ I thought maybe I could just say if $$f$$ is in the space so is $$(a-b)f$$ so I'd be done but I'm not convinced thats correct.

$$|\int_a^bk(x,y)(f(y)-s(y))dy|\leq\int_a^b|k(x,y)|sup_{y\in[a,b]}|f(y)-s(y)|dy=\|f-s\|\int_a^b|k(x,y)|dy$$.