The question is as follows:
The circle $S^1$ is defined by $\mathbb R$/$\mathbb Z$, where $\mathbb Z$ is viewed as an additive subgroup of $\mathbb R$ and where $\mathbb R$ acts on itself by addition. Prove that $S^1$ is a manifold. Prove that if $X$ is a connected, compact, one–dimensional manifold, then $X$ is homeomorphic to $S^1$.
I only have a very basic understanding of manifolds and am struggling on how to approach this problem. If it is possible to explain the logic/rational behind the steps in the proof that would be greatly appreciated.
Showing that $S^1$ is a manifold requires showing that it is Hausdorff (which is okay because $\Bbb Z$ is discrete in $\Bbb R$), second-countable (which is okay because $\Bbb R$ is second-countable and the orbits under $\mathbb Z$ are discrete, so projection is a local homeomorphism), and locally euclidean (again, because projection is a local homeomorphism).
The more interesting part is to show that all connected compact $1$-manifolds are homeomorphic, but I cannot do this in an answer here, maybe you should check out the classification of $1$-manifolds in Milnor's Topology from the Differentiable Viewpoint. The basic idea is as follows: Cover the connected compact $1$-manifold $M$ by two connected charts whose intersection is disconnected. Then their pre-image in $\mathbb R$, arranged in a nice way, must "wrap up", so some points will be identified. Iterating this, we get that $M$ is homeomorphic to $\mathbb R/ \mathbb Z$.