Proving a given set is a submanifold Let $S \subseteq \mathbb R^n$. I have been faced with showing that $S$ is a submanifold and I have some ideas but I want to get the complete picture.

(Main) Question 1: What methods are there to prove that something is a (sub-)manifold? 
  (I am putting ''sub'' into brackets as I believe showing the
  properties of a manifold is sufficient as a subset which is a manifold
  is, of course, a submanifold)

My ideas:
(1) There is a theorem about smooth maps and it goes like this:
Let $f: M \to N$ be smooth and let $y \in N$ be a regular value. Then $f^{-1}(y)$ is a submanifold of $M$.
So, given some set, if I can define a suitable $f$ then I can apply this theorem. But it seems to me that having to restrict to inverse images of regular values greatly restricts what sets we can prove are manifolds. 

(Sub-)Question 2: Am I missing something or is this theorem indeed kind of limited in
  its applications?

(2) The next method that I could think of was to produce some (smooth) charts. 
Concretely, say we want to prove that $S^1$ is a smooth $1$-manifold then we can note that $\{(U,\phi), (V,\psi), (W, \eta)\}$ is a smooth atlas where $U = (-\pi,\pi), \phi: U \to S^1, \phi(x) = e^{2 \pi i}$, $V = (-\pi,\pi), \psi: U \to S^1, \psi(x) = e^{2 \pi i + {3\pi \over 4}}$ and $W = (-\pi,\pi), \eta: U \to S^1, \eta(x) = e^{2 \pi i + {3\pi \over 2}} $.
The problem I have with this is that I think for it to be a smooth structure the atlas would have to be maximal. But I don't know how to prove that this atlas is maximal (and maybe it isn't).

(Sub-)Question 3: Can someone elaborate on this method of explicit
  specification of the atlas and about maximal atlases? Also, again it
  seems to me that this method is quite limited as it does not seem
  practical to write down an entire atlas in general (sure, $S^1$ was
  not much work but I suppose it is impossible for worse sets).

 A: If $M$ is a closed codimension-1 submanifold of $\mathbb R^n$, smoothly embedded, then the first method is in fact completely general, in the sense that if you have such a manifold, you can compute the signed distance to $M$ and call this $f$; $0$ is then a regular value, and $f^{-1}(0) = M$. 
For things like the (open) Mobius band, however, that's not enough, and a patch-based method is really needed. Fortunately, for almost every such (smooth) manifold, for every point there's a neighborhood -- often small, alas -- with the property that projection from that neighborhood to some coordinate plane (the $xy$-plane, the $yz$-plane, etc.) is in fact a diffeomorphism, and hence is what you need to define a local chart. 
These two ideas suffice to let you prove a remarkable variety of manifolds are indeed manifolds, including pretty much all the classical groups, for instance, plus spheres, balls, tori, products of manifolds, etc. 
Let me amplify on that second approach, by applying it to the circle. One "patch" for the circle would be the (open) upper half-circle, with the map $(x, y) \mapsto x$; others would be the right half-circle, the left half-circle, and the bottom half-circle. The transformation from the upper to the right half-circle domains ends up being a map from $(0, 1)$ on the $x$-axis to $(0,1)$ on the $Y$-axis; the inverse of the first map sends $x$ to $(x, \sqrt{1 - x^2})$, and the second map sends this to the $y$-coordinate, i.e, to $\sqrt{1 - x^2}$. The composition of those two is a nice map from the open interval to itself. 
As for the matter of "maximal" atlases -- that's a technicality, designed to make certain proofs easier. Once you have an atlas that covers your whole manifold, there's a unique maximal atlas associated to it, so you've done enough. No one ever writes down maximal atlases, because they're rather large. :)
