TITLE: Why are there no wild arcs in the plane?
QUESTION [33 upvotes]: On math.stackexchange it was asked whether all arcs in the plane are ambient-isotopic. I suggested that one could prove this by appealing to the Schönflies theorem, which you can do as long as you can extend your Jordan arc to a Jordan curve. That is, as long as you can extend an embedding of an interval to an embedding of a circle. However, I have to admit I don't readily see how to do this, and there are examples which show this is a subtle question. E.g. you can take an arc which spirals infinitely into a point. So my question is how one can show an embedding of an arc into $\mathbb R^2$ extends to an embedding of a circle into $\mathbb R^2$, or, failing that, if someone knows some other proof that all planar Jordan arcs are standard.
Edit: I want to highlight Bill Thurston's elegant answer (in a comment) to the question of whether you can extend an arc to a circle, even though I accepted his other answer using Caratheodory's theorem. Namely, assume your arc runs from $0$ to $\infty$ in $\mathbb C\cup\{\infty\}$. Then take the pre-image under the double branched cover $z\mapsto z^2$. The original arc can be identified with one of its two preimages, while the other preimage fills it in to a circle. Then one can apply the Schönflies theorem.

REPLY [2 votes]: The desired result appears in M.H.A. Newman's classic book, Elements of the Topology of Plane Sets of Points (2nd ed., 1951), as Theorem 14.5 in Chapter VI, on pg. 164:
Theorem 14.5: Every simple arc in $X^2$ is an arc of a simple closed curve in $X^2$.
$X^2$ is Newman's notation for a space that is either the "open" plane, $R^2$, or the "closed" plane, $R^2 \cup \{\infty\}$.
N.B. The proof of Theorem 14.5 does not involve Schoenflies' Theorem. Also, Newman's definition of an arc is such that it has ends.