real analysis - Help Debugging a Bogus Proof

We want to prove the standard fact that a smooth function $u :R^2 \to R$ with $\nabla u = 0$ everywhere in some connected open set $\Omega$ is constant in that set.

I'm comfortable with the usual proof, but I also thought it could be proved the following way:

$$\nabla u = 0 \space\iff\space du = 0$$

$u$ also satisfies Laplace's equation($\nabla \cdot\nabla u = 0$), and hence has the mean value property.

Let $x_0$ be any point of $\Omega$(distinct from the origin!). For any disk $\Delta$ about $x_0$ of radius $r$ sufficiently small, $$u(x_o) = \frac {1} {2\pi} \int_{\partial\Delta}u\space d\theta$$

But, by Stokes' theorem $$\int_{\partial\Delta}u \space d \theta = \iint_{\Delta}du\wedge d \theta = 0 \tag1$$ (as $d^2 = 0$, and $du = 0$).

This is not the correct conclusion, as $u$ could be equal to any constant.

I think $(1)$ must be the problematic line, but I can't see why, as $$d(u \space d\theta) = du\wedge d\theta - u \wedge d^2 \theta$$ .