How do you prove Reidemeister’s theorem for wild knots? Reidemeister’s theorem states that any two diagrams for the same knot are related by the three Reidemeister moves.  Now the way Reidemeister, Alexander, and Briggs originally proved it is by showing how the Reidemeister moves are projections of triangle moves (AKA delta moves) of polygonal knots.  But only tame knots are equivalent to polygonal knots.  So my question is, how do you prove Reidemeister’s theorem for wild knots?
Is there an elementary proof for wild knots, maybe reducing the problem to the tame knot case somehow?
 A: First of all, a knot in $R^3$ is tame if it is isotopic to a polygonal knot. This is not the same thing as to say that the knot itself is polygonal. I checked a number of books on knot theory (Crowell and Fox; Lickorish; Kauffman; Manturov; Murasugi) and from the beginning, after discussing briefly wild knots, they all assume knots to be either $C^1$-smooth or PL (piecewise-linear or polygonal). Rolfsen's book is the only exception: He first discusses topological knots and only after page 60 he imposes the PL condition when introducing "knot pictures." Literally, nobody discusses knot diagrams for topological knots, so I am not sure what is your source. (As a rule of thumb, one should not trust Wikipedia to have enough rigor on these matters.) 
Thus, it is no surprise that, by the time of the discussion of Reidemeister Moves (RMs), nobody mentions these polygonality/smoothness conditions: These are assumed by default. 
Here is a formal definition of knot diagrams for topological knots in $R^3$; I am following Manturov's (V. Manturov, "Knot Theory") treatment as the most careful, but I extend it to the case of topological knots. 
Let $f: S^1\to R^3$ be a topological embedding (a topological knot $K$) and let $P\subset R^3$ be a cooriented plane: Every line orthogonal to $P$ comes equipped with an orientation.   Let $\pi=\pi_P: R^3\to P$ denote the orthogonal projection. Set $g:=  \pi\circ f: S^1\to P$. 
We will say that $K$ is regular with respect to $P$ if the following conditions are met:


*

*For every $p\in P$, $g^{-1}(p)$ consists of at most two points. Define the subset $T\subset S^1$ consisting of those $t\in S^1$ for which $g^{-1}(p)$ has cardinality $2$. This is the subset of "double points" of the self-intersection of the map $g$. 

*The subset $T$ is finite. In particular, for every $t\in T$ there exists a small arc $\alpha_t\subset S^1$, a neighborhood of $t$, such that  $g$ restricted to $\alpha_t$ is 1-1. 

*For any two distinct $t_1, t_2\in T$, the restrictions $g|_{\alpha_{t_1}}, g|_{\alpha_{t_2}}$ are topologically transversal to each other. (In the context of polygonal knots this is usually expressed by the requirement that no vertex of a polygonal knot projects to a double point of the knot diagram.) I will not define topological transversality (intuitively, one topological arc in $R^2$ lies on "both sides" of the other), but it is critical for the definition of RMs on knot diagrams. 
A key observation is that if $f(S^1)$ is smooth or polygonal, then $K$ is regular with respect to a "generic" plane $P$. This, however, is quite false for general topological knots (even for trivial ones!). A good example to think about is an unknot whose projection $g$ is a space-filling curve in the plane $P$. 
The following lemma is a pleasant exercise in knot theory:
Lemma. If $K$ is a knot regular with respect to a plane $P$, then $K$ is tame. 
I skip the proof and note only that for the proof one needs only the assumptions 1 and 2; moreover, 1 can be replaced by the weaker condition that   $g^{-1}(p)$ is finite for every $p\in P$. 
Now, given this, define the knot diagram $D_{K,P}$ of the knot $K$ regular with respect to $P$ as the following data:
(a) The composition $g=\pi\circ f: S^1\to P$.
(b) A function $c: T\to \{o, u\}$ with values in the 2-element set $\{o, u\}$, which records the "over" and "under" crossings of the knot $K$ with respect to $P$. For each $t_1\in T$, let $g^{-1}(g(t_1))=\{t_1, t_2\}$. 
If $f(t_1)> f(t_2)$ on the oriented line through these two points, then $c(t_1)=o$ (the point $f(t_1)$ is "over"    $f(t_2)$) and $c(t_1)=u$ otherwise. In particular, $c(\{t_1, t_2\})=\{o, u\}$. 
This over/under information is typically recorded by making a little break at $g(t_2)$ when drawing the "under" arc $g(\alpha_{t_2})$, whenever $c(t_2)=u$.  
Thus, a knot diagram is much more than just the composition $g: S^1\to P$. The key property of a knot diagram is that, unlike $g$ alone, the diagram $D_{K,P}$  determines the isotopy class of $K$ in $R^3$. 
One can attempt to define analogues of knot diagrams for general topological knots (without the regularity assumption: Recording the total  ordering of the point-preimages under $g$, determined by orientations on lines orthogonal to $P$). I do not know what would this be good for and if one can prove anything using it. I also do not know how to define RMs without the regularity assumption. Saying that they are "the same" as in the polygonal case is meaningless. Thus, your request for a proof of Reidemeister’s theorem for wild knots is hopeless, since it is unclear even how to define knot diagrams and RMs in this context. (By Lemma above, regularity implies tameness.) 
