Proof that exact form are path independent seems to imply the same for merely closed forms A singular $k$-cube on some set $A \subseteq \mathbb R^n$ is a continuous map $c : [0,1]^k \to A$. Consider the following exercise:

Let $c_1, c_2$ be singular $1$-cubes in $\mathbb R^2$ with $c_1(0) = c_2(0)$ and $c_1(1) = c_2(1)$. Show that $\int_{c_1} \omega = \int_{c_2} \omega$ if $\omega$ is exact. Give a counter-example on $\mathbb R^2 - 0$ if $\omega$ is merely closed.

Here $\omega$ denotes a differential $1$-form on $\mathbb R^2$, i.e. for each $p \in \mathbb R^2$ we have that $\omega(p)$ is a linear map on $\mathbb R^2_p$, i.e. the tangent space at the point $p$. This is exercise 4-32 (a) from Spivak: Calculus on Manifolds (page 105), and a solution could be found here. The solution goes like this: 
First it is shown that there exists a $2$-cube $c : [0,2]\to \mathbb R^2$ such that $\partial c = c_1 - c_2 + c_1(1) - c_1(0)$, and then using that on $1$-cubes the integral vanishes and Stokes theorem:

Suppose $\omega$ is exact, hence closed. Then by Stokes Theorem we have $\int_{c_1 - c_2} \omega = \int_{\partial c} \omega = \int_c d\omega = \int_c 0 = 0$ (since $d\omega = 0$ as it is closed), and so $\int_{c_1} \omega = \int_{c_2}\omega$.

So as I see it just uses the fact that $\omega$ is closed, but not the more stronger property of exactness. So this proof would also work if $\omega$ is merely closed. What have I overlooked here?

You are somehow right: the proof would work if you only knew that $\omega$ were closed, provided that $\omega$ were defined not just on $\partial C$, but on all $C$. But in that case, all closed forms on $C$ are exact. 
That is not the case on $\mathbb{R}^2 \setminus \{ 0\}$. If you pick $C_1$ and $C_2$ to bound a region $C$ that does not contain $0$, then it is true that $\int_{C_1}\omega = \int_{C_2}\omega$ for every closed form $\omega$. However, on such a region, all closed forms are exact. The problem occurs when the region $C$ contains $0$ - in that situation $d\omega$ is not defined on $C$.
