I am asked to show that $H^{1}_{c}(\mathbb{R})\cong\mathbb{R}$.
In coordinates, a compactly supported 1-form on $\mathbb{R}$ can be written as $\omega=f(x)\mathrm{d}x$, where $f(x)$ is a compactly supported function. Lets assume further that $\mathrm{supp}(f)\subset B_{R}(0)$ for a $R>0$, which is possible, because $f$ is compactly supported.
$\omega$ is obviously closed, because $\mathrm{d}\omega=\frac{\partial f(x)}{\partial x}\mathrm{d}x\wedge \mathrm{d}x=0$. If we define now the primitive of $f$ as $F(x):=\int_{-2R}^{x}f(x)\mathrm{d}x$, then we see that $\omega=\mathrm{d}F$ and that $F$ is compactly supported.....Hence $\omega$ is exact....But wouldn't this imply that all compactly supported closed 1-forms are exact and therefore $H^{1}_{c}(\mathbb{R})\cong 0$? Where is the mistake in my "proof"?
Your function $F$ is not necessarily compactly supported. Indeed, for all $t>R$, $F(t)=\int_{\mathbb{R}}f(x)\,dx$ which may be nonzero. So, your argument only shows that $\omega$ is a coboundary if its integral over all of $\mathbb{R}$ is $0$. In fact, the map $H^1_c(\mathbb{R})\to\mathbb{R}$ sending $[\omega]$ to $\int_\mathbb{R}\omega$ is an isomorphism (see if you can prove that!).