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The smallest atlas for the $S^1=\{(x,y)\in \Bbb R^2|\,x^2+y^2=1\}$ must contain two charts. How to prove it?

My route is first to prove that it can only be homeomorphic to $ \Bbb R^1$ by invariant of domain (dimension). Then it restricts on mapping from $S^1$ into $\Bbb R$.

Secondly, I want to show such a homeomorphism cannot exist. But I got stuck, since I do not know how to partition a circle and I noticed that the topological structure on the circle can be defined by those arcs on the circle with one end be filled one end be empty. Then for the topology structure one should have a homeomophism eventually map$: (a,b]\rightarrow (c,d)$ which is not true. Then I do not know what is wrong.

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    Hint: the circle is compact – 2017-01-30
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    @AlessandroCodenotti Brilliant! It could be the simplest proof that a circle cannot have atlas with one chart. – 2017-01-31
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    Sam's proof is also correct clearly, but the approach I suggested generalizes to all compact manifolds immediately – 2017-01-31

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If $U$ is any (nonempty) open subset of $\mathbb{R}$, then $U$ minus any point is disconnected. However, $S^1$ minus a point is always still connected.