2 added 1294 characters in body; added 54 characters in body

Here are some extra remarks about the argument. First, this combinatorial principle. Another illustration of the same principle is the fact that a graph is $k$-colorable if and only if every finite subgraph is $k$-colorable. If the graph only has countably many vertices, then there is a standard proof by induction; it is only the uncountable case that requires something fancier such as the ultrafilter lemma or Tychonoff's theorem. In fact, the case that's needed is almost the same as the $k=2$ case of colorability. This case, and the orientability argument, is even easier than the general case because the local coloring or orientation is essentially unique.

Second, cleaning up the atlas so that every chart is an interval and a non-empty intersection of any two charts is an interval. The first condition is sometimes part of the definition of an atlas. But if not, every open set in $\mathbb{R}$ is a countable union of intervals (or every open set in $\mathbb{R}^n$ is a countable union of balls) and you can just make them separate charts. As for the second condition, using the intermediate value theorem and the Hausdorff condition, two interval charts can only intersect at one end or at both ends. If they intersect at both ends, then all of the other charts are redundant and the manifold is a circle.

1

I think that it's easiest to model an orientable manifold as one whose gluing maps are in $\mathrm{Diff}^+$. Actually the following proof also works in the topological category.

You can assume that all of the charts of the 1-manifold are open intervals, and that any two charts also intersect in an interval. Then, once you orient one of the intervals, the orientation spreads to its neighbors. Now, there is a principle in combinatorics that the orientations will all be consistent unless there is a finite obstruction. What would this obstruction look like? Using the Hausdorff condition, and throwing away redundant charts, you can clean up any finite collection of charts until you either have a sequence of charts or a cyclic sequence of charts chained together at the ends. Then it's clear in either case that there is no finite obstruction.

Note that there are non-Hausdorff 1-manifolds that are not orientable; the Hausdorff condition is thus essential to the proof.