Hi ZaidAlyafey,
I'm assuming we are taking $$f$$ to be meromorphic on an open subset, say $$U$$, of $$\mathbb{C}$$ that contains $$c.$$ I also assume the circular arcs we are considering contain the endpoints of the arc (i.e. are compact). That said, we can prove the claim by looking at the Laurent series expansion for $$f$$.
Since $$z=c$$ is a simple pole of $$f,$$ we know that the Laurent series expansion for $$f$$ is of the form
$$f(z)=\frac{Res(f;c)}{z-c}+\sum_{n=0}^{\infty}a_{n}(z-c)^{n}\qquad (*),$$
where the above holds on an annulus of inner radius $$\rho$$ and outer radius $$R.$$ Now
$$\rho = \limsup_{n\rightarrow\infty}|a_{-n}|^{1/n}=0,$$
and since $$f$$ is meromorphic on $$U,$$ $$R>0;$$ the fact that $$\rho=0$$ is what allows us to let $$r\rightarrow 0^{+}.$$ We must note that $$(*)$$ converges uniformly on compact subsets of our annular region (
Laurent series - Wikipedia, the free encyclopedia).
Since, for small enough $$r,$$ $$C_{r}$$ is a compact subset of our annular region, $$(*)$$ converges uniformly on $$C_{r}.$$ Using the uniform convergence to justify interchanging integral and sum we have
$$\int_{C_{r}}f(z)dz=Res(f;c)\int_{C_{r}}\frac{1}{z-c}dz+\sum_{n=0}^{\infty}a_{n}\int_{C_{r}}(z-c)^{n}dz.$$
Parameterizing $$C_{r}$$ via $$z=c+re^{i\theta}$$ and integrating gives
$$\int_{C_{r}}f(z)dz=i(\theta_{2}-\theta_{1})Res(f;c)+ir\sum_{n=0}^{\infty}a_{n}\int_{\theta_{1}}^{\theta_{2}}r^{n}e^{i(n+1)\theta}d \theta$$
The sum on the right exists, because the integral on the left exists. Hence, taking the limit we obtain
$$\lim_{r\rightarrow 0^{+}}\int_{C_{r}}f(z)dz=i(\theta_{2}-\theta_{1})Res(f;c),$$
as desired
