let C be a smooth simple closed curve in the plane and n be a unit vector normal to C.
prove that the integral of n.ds over C is 0.
You cannot compute a line integral over a vector field.
Therefore the integral of n.ds over just makes no sense.
I think you might mean the following:
Define the vector-field $\displaystyle \bold{F}: \mathbb{R}^2 \to \mathbb{R}^2$ by $\displaystyle \bold{F}(x,y) = (1,1)$.
From here we see that $\displaystyle \text{div}(\bold{F}) = 0$ and $\displaystyle \bold{F}\cdot \bold{n} = n_1 + n_2$ (sum of components of the normal vector).
Thus, by divergence theorem,
$\displaystyle 0=\iint_S \text{div}(\bold{F}) dA = \oint_C (n_1+n_2) ds$