1-manifold is orientable I am trying to classify all compact 1-manifolds. I believe I can do it once I can show every 1-manifold is orientable. I have tried to show prove this a bunch of ways, but I can't get anywhere.
Please help,
Note, I am NOT assuming that I already know the only such manifolds are [0,1] or $S^1$. This is my end goal.  
 A: If you've already classified orientable $1$-manifolds, then you know that the only connected ones (without boundary) are $\mathbb R$ and $\mathbb S^1$. Now suppose $M$ is a connected, nonorientable $1$-manifold, and let $\pi\colon \widetilde M\to M$ be its universal covering.  Then $\widetilde M$ is orientable and simply connected, and therefore homeomorphic to $\mathbb R$, and $M$ is homeomorphic to a quotient of $\mathbb R$ by a free group action that does not preserve orientation.  The last step is to show that every orientation-reversing homeomorphism $f\colon \mathbb R\to \mathbb R$ has a fixed point, which yields a contradiction.  Thus every $1$-manifold is orientable.
For a $1$-manifold with (nonempty) boundary, you can apply the above argument to the double of $M$ (the quotient of two disjoint copies of $M$ obtained by identifying each boundary point in one copy with the corresponding boundary point in the other).
A: There are only two connected, compact 1-manifolds up to homeomorphism.  These are $[0,1]$ and $\mathbb{S}^1$ (the circle).  These will (very likely) be trivially orientable given your definition of orientability.
A: If I understand your question correctly, I guess you have proven that all the connected orientable 1-manifolds are $[0,1]$ and $S^1$. Now suppose that $X$ is connected but not orientable. Then $X$ has a universal covering space $\tilde X$ which is still a 1-manifold but is orientable. Both cases $\tilde X = [0,1]$ or $S^1$ give a contradiction.
A: You could also try to do this. Suppose $X$ is a 1-manifold and $f:S^1 \to X$ is an orientation-reversing path. If $f$ is a smooth mapping without critical points, then this is a contradiction (because you can take the image under $df$ of the non-vanishing vector field on $S^1$, thus showing that path $f$ is not orientation-reversing).
Now, for a general path $f$ (that can be non-smooth, or can have critical points) it should be possible to prove that $f$ is homotopic to a path $g: S^1 \to X$ that is smooth and has no critical points. If this is shown, then $g$ must also be orientation-reversing (since it is homotopic to $f$), and we arrive at the same contradiction as before.
So, it all hinges on this lemma: if a path $f: S^1 \to X$ isn't contractible, then it is homotopic to a path $g: S^1 \to X$ that is smooth and has no critical points. Note that non-contractibility is necessary here, but we have it (because contractible paths can't be orientation-reversing anyway).
The smoothness part isn't that hard, after all each path can be approximated with a smooth one. So we may assume that $f$ is smooth from the start. The problem is that $f$ may have critical points, i.e. the path can stop at several points. At this point I'm afraid some technical work is necessary to show that these stopping points can be eliminated, either by "smoothing" (if we stop and then continue going in the same direction), or by retracting dead ends (if we stop and go backwards). This can be formalized, but it's not pretty. The way that comes to mind is to introduce a Riemannian metric on $X$ (which always exists) and look at piecewise smooth paths in which each smooth piece has fixed speed w.r.t. the chosen metric. In this setting it is possible to formalize the idea of removing dead ends and obtain the desired path without critical points.
There's some unpleasant technical work, I agree. But on the other hand all the machinery is visible and graspable.
