Line bundles of the circle Up to isomorphism, I think there exist only two line bundles of the circle: the trivial bundle (diffeomorphic to a cylinder) and a bundle that looks like to a Möbius band. Although it seems obvious geometrically I did not find a good argument to justify it. Do you have an idea?
 A: I'll show how the powerful (but sophisticated) theory of sheaves permits one to classify  real  line bundles on a paracompact topological space.
This will delight algebraic geometers and maybe motivate others to learn that theory.
Let $\mathcal C^*$ (resp. $\mathcal C^*_+)$ denote the sheaf of continuous nowhere zero functions (resp. sheaf of continuous positive functions) on a paracompact space $X$.
The exact sequence  $ 0\to   \mathcal C^*_+\to \mathcal C^*\stackrel {\text {sign}}\to \mathbb Z/2\mathbb Z\to0$ gives rise to a long exact sequence in cohomology of which a fragment is $$   \cdots \to H^1(X,\mathcal C^*_+)   \to  H^1(X,\mathcal C^*)  \to   H^1(X,\mathbb Z/2\mathbb Z)\to   H^2(X,\mathcal C^*_+) \to\cdots$$
Now we have an isomorphism of sheaves  $\mathcal C^*_+ \stackrel {\log} {\cong}\mathcal C$ and thus $\mathcal C^*_+$ is acyclic because $\mathcal C$ is acyclic (since it is a fine sheaf by paracompactness of $X$).
In particular  $H^1(X,\mathcal C^*_+) = H^2(X,\mathcal C^*_+)=0$ 
so  that the above cohomological fragment reduces to $   0  \to  H^1(X,\mathcal C^*)  \to   H^1(X,\mathbb Z/2\mathbb Z)\to   0      $ and since $H^1(X,\mathcal C^*)$ classifies line bundles on $X$ we get the result that line bundles on $X$ are classified by $H^1(X,\mathbb Z/2\mathbb Z)$.       
In the differential geometry category the analogous result holds with $\mathcal C$ replaced by $\mathcal C^\infty$.
This yields the astonishing result that on a manifold each continuous line bundle has one and only one differential structure (up to isomorphism).  
Finally, for the circle $H^1(S^1,\mathbb Z/2\mathbb Z)=\mathbb Z/2\mathbb Z$ and this proves your result.  
Remark
The main reason I am posting this proof is for my record: I have never seen it in a reference  and I want to be able to retrieve it in the probable case that I forget it! 
A: You can show that the two line bundles have different first Stiefel-Whitney classes.
Added later: This shows that the two bundles are not isomorphic, not that these two bundles are the only ones (up to isomorphism).
A: Line bundles are classified by the first Čech cohomology with coefficients in $\text{GL}^1(\mathbf R)$. By normalizing we can actually use coefficients in $\text{O}^1(\mathbf R)=\pm 1$. By using the usual covering of the circle, one sees immediately that this cohomology group is cyclic of order $2$, generated by the class of the Möbius strip.
A: Here is the sketch of an elementary proof:
Let $E \overset{\pi}{\longrightarrow} \mathbb{S}^1$ be a line bundle and $p : \mathbb{R} \to \mathbb{S}^1$ be the usual universal covering. 


*

*Let $F= \coprod\limits_{x \in \mathbb{R}} E_{p(x)}$ be a line bundle $F \overset{q}{\longrightarrow} \mathbb{R}$ defined by the local trivializations $\varphi= (p_{|V}^{-1} \times \operatorname{Id}) \circ \phi \circ (p \times \operatorname{Id})$ where $\phi$ is a local trivialization for $E \overset{\pi}{\longrightarrow} \mathbb{S}^1$ and $V$ is an elementary neighborhood of the overing $\mathbb{R} \to \mathbb{S}^1$. Then $p \times \operatorname{Id} : F \to E$ is a morphism.

*Because $\mathbb{R}$ is contractible, every vector bundle on it is trivial (see here). Let $\psi : F \to \mathbb{R} \times \mathbb{R}$ be a global trivialization.

*Notice that $(p \circ \operatorname{Id}) \circ \psi$ defines a surjective smooth map between $[0,2\pi] \times \mathbb{R}$ and $E$, injective on $[0,2\pi) \times \mathbb{R}$. Moreover, it induces linear isomorphisms $f_1 : \{2\pi\} \times \mathbb{R} \to E_0$ and $f_0 : \{0\} \times \mathbb{R} \to E_0$. For convenience, let $f= f_2^{-1} \circ f_1$.

*We deduce the diffeomorphism $$E \simeq [0,2\pi] \times \mathbb{R} / \{ f(2\pi,x) \simeq (0,x) \}.$$ Without loss of generality, we may normalize $f$ so that $f= \operatorname{Id}$ or $f=- \operatorname{Id}$. Therefore, two cases happen: either $$E \simeq [0,2\pi] \times \mathbb{R} / \{(0,x) \sim (2\pi,x) \} \simeq \mathcal{C},$$ or $$E \simeq [0,2\pi] \times \mathbb{R} / \{(0,x) \sim (2\pi,-x) \} \simeq \mathbb{M}.$$

