Let $\mathbb{T}:=\mathbb{R}/\mathbb{Z}$ and $\pi:\mathbb{R}\to\mathbb{T}$ the natural projection. Show that there exists a closed, non-exact $1$-form $\theta\in\Omega^1(\mathbb{T})$ such that $\pi^*\theta=dx$.
Here is what I've done: take the submanifold $(0,1)\subset\mathbb{R}$ and notice that $\left.\pi\right|_{(0,1)}:(0,1)\to\pi((0,1))$ is a diffeomorphism. Taking $dx\in\Omega^1((0,1))$, define $\theta:=(\left.\pi\right|_{(0,1)}^{-1})^*(dx)$. In deed we have $\pi^*\theta=dx$ and $\theta$ is closed, since $d\theta=d(\left.\pi\right|_{(0,1)}^{-1})^*(dx)=(\left.\pi\right|_{(0,1)}^{-1})^*(d^2x)=0$. Also, $\theta$ cannot be exact, because if there is some $\sigma\in\Omega^0(\mathbb{T})$ with $\theta=d\sigma$, then:
$$\int_{\pi((0,1))} \theta=\int_{(0,1)}d\sigma=\int_{\partial(0,1)=\emptyset}\sigma=0$$
which is absurd, since $\int_{\pi((0,1))} \theta=\int_{(0,1)}\pi^*\theta=\int_0^1 dx=1$.
I'm not sure about this solution since, rigorously speaking, $\theta$ is a $1$-form in $\pi((0,1))$, not in the whole $\mathbb{T}$, which I've implicitly ignored. So the claim $p^*\theta=dx$ is not very well formalized.
Does this solution work? Why? How can it be better formalized?
As you have observed, $\pi | (0,1)$ does not cover the whole $\mathbb{T}$, so your $1$-form $\theta$ is not defined on the whole of $\mathbb{T}$. So let's not call it $\theta$, let's call it instead $\theta_1$, and let's remember that $\theta_1$ is only a 1-form on $\mathbb{T} - \{\pi(0)\}$.
But now you can repeat the process with $\pi | (-1/2,1/2)$. You'll get a 1-form $\theta_2$ defined on $\mathbb{T} - \{\pi(1/2)\}$.
Your remaining job is to check that on the overlapping set $\mathbb{T} - \{\pi(0),\pi(1/2)\}$, the restrictions of $\theta_1$ and $\theta_2$ are equal. So, you can paste them together to obtain the desired 1-form $\theta$ defined on all of $\mathbb{T}$.