Given $f(z)$ is analytic everywhere inside and on $C$. Let $C$ and $D$ be two closed contours, $D$ lying completely within $C$ (both the same orientation) , and let $a$ be a point between $C$ and $D$. Show that:
$$ f(a) = \frac{1}{2\pi i}\int_{C}\frac{f(z)}{z-a}dz - \frac{1}{2\pi i}\int_D \frac{f(z)}{z-a}dz. $$
This proof involves multiply connected theorem. I have tried to prove the R.T.P constructing another simple closed contour $E$ between $C$ and $D$ of the opposite orientation but have had little success; any help is appreciated.
Since $a$ is "outside" the curve $D$, the function $\dfrac{f(z)}{z-a}$ is Holomorphic on and inside $D$ and, therefore, $\int_D \dfrac{f(z)}{z-a} dz = 0$. So, your results just amounts to Cauchy's integral formula on $C$. $$ \int_C \dfrac{f(z)}{z-a}dz - \underbrace{\int_D \dfrac{f(z)}{z-a}dz}_{=0} = \int_C \dfrac{f(z)}{z-a}dz = 2 \pi i f(a).$$
This leads to the result you want to prove (you are missing an $i$ ).