Is there a fixed point theorem I could use to solve this problem? let $E = C([0,1]),\,\,$ $K : E \to E, \,\,
(Kf)(x) = \int_0^1K(x,y)f(y)dy$
also $\|K\| \leq a < 1$
I want to prove that there for $g \in E$ there exists a unique $f_g \in E$ that satisfies the following equation : 
$f_g + Kf_g = g$
which is equivalent to showing that $T : E \to E,\,\,T(f) = g-Kf$ has a fixed point.
with what I have in hands I feel like there must be some theorem I'm missing.
any help will be greatly appreciated !  

This is not a fixed point theorem, but it is well-known that if $T:E\to E$ is a bounded linear operator with $\|I-T\|<1,$ then $T$ has a bounded inverse
$$
T^{-1}=\sum_{k=0}^\infty (I-T)^k.
$$ In your case, since $\|I-(I+K)\|<1$, we have
$$
f_g = (I+K)^{-1}g,\quad \|f_g\|\leq \|(I+K)^{-1}\|\|g\|.
$$
