How to prove that any discrete set without limit points in the plane maps to the integers via ambient isotopy? Which extension theorem to cite? Questions. 
(0) What are efficient reasons for: any discrete subset of $\mathbb{R}^2$ without limit points can be mapped by a plane isotopy to the pointset $\{ (x,0)\colon x\in\mathbb{Z}\}\subseteq\mathbb{R}^2$?
(1) Is it true or false that: any countably-infinite subset of $\mathbb{S}^2$ with precisely one limit point can be mapped by an ambient isotopy of the sphere $\mathbb{S}^2$ to the pointset $\{ (\cos(\frac{2\pi}{r}),\sin(\frac{2\pi}{r}),0)\colon r\in\mathbb{Z}, r>0 \}\subseteq\mathbb{S}^2$?
Remarks.


*

*The questions thelselves should be complete without further explanations. 

*One obvious sort-of-conceptual approach to (0) would be to say something like '$\mathbb{R}^2\setminus S$ is homeomorphic to 'the' surface with countably-many and discretely-spaced punctures.", yet I neither found a citable reference useful to answer the question proper, nor found it useful to bring in the concept of 'surface' here wwhen explaining/justifying this to someone else. 

*I think I could work out a proof myself, yet I would prefer a precise citable reference in the published literature. This is such an intuitively evident and elementary statement that it seems likely that it occurs several times in the literature. 

*To me, the obvious approach to (0) seems this. Given any discrete set $S\subseteq\mathbb{R}^2$, $S$ is necessarily countable, so there is an bijection $p\colon I\rightarrow S$, $i\mapsto p_i$, with $I$ an at most countably infinite set. 
We may assume w.l.o.g. that $S$ is countably infinite, and that $I=\mathbb{Z}$. 
One then can of course define the map 
$\vartheta\colon S\times [0,1]\rightarrow\mathbb{R}^2$ given by 
for all $i\in\mathbb{Z}$, for all $t\in[0,1]$, $\qquad\theta(p_i,t) = (1-t)\cdot p_i + t\cdot i$
and one now needs a precise reference to a suitable extension theorem which would guarantee an extension of $\vartheta$ to a plane isotopy $\theta\colon\mathbb{R}^2\times[0,1]\rightarrow\mathbb{R}^2$. In this sense, a variant question whose answer would be:


*

*What is a right extension theorem to extend the above $\theta$? For obvious reasons, the right direction seem to be known statements about continous maps of the form $[0,1]^{\omega}\rightarrow\mathbb{R}^n$. 
What do you consider the most standard theorem and reference to cite here?

*While plane isotopy is a very usual term, here is a definition, copied from this MO thread, a question which is related to the motivation for this question (an open problem about infinite planar graphs)


*

*Let 



$\eta_S\colon \mathbb{R}\rightarrow\mathbb{R}^2$,
and 
$\eta_R\colon \mathbb{R}\rightarrow\mathbb{R}^2$,
be arbitrary set-maps (for the purposes of the discrete sets, we could of course restrict to piecewise continuous, or even piecewise differentiable maps)
Then a plane isotopy from $\eta_S$ to $\eta_R$ is any continuous set-map 
$\theta\colon \mathbb{R}^2\times[0,1]\rightarrow\mathbb{R}^2$
satisfying the three axioms
(A.0) for all $v\in\mathbb{R}^2$, $v=\theta(v,0)$,
(A.1) for all $t\in[0,1]$, $v\mapsto\theta(v,t)$ defines a homeomorphism $\mathbb{R}^2\rightarrow\mathbb{R}^2$,
(A.2) $(v\mapsto \theta(v,1))\circ\eta_S = \eta_R$, equal as set-maps.


*

*I would appreciate both references to the question proper, and references to extension statements appropriate to the evident 'write-down-the-isotopy' approach. It should be easily be provable that one can ' extend $\vartheta$ 'piecewise-linearly' to a plane isotopy $\mathbb{R}^2\times[0,1]\rightarrow\mathbb{R}^2$ ' , yet there should be a better approach than try to do this explicitly by hand.

*Are there relevant references on (0) and (1) from the published literature on configuration spaces?
 A: You need to use the 1963 paper by I.Richards "On the classification of noncompact surfaces". It takes some time to get through the terminology, but the "ideal boundary" he is talking about (or "the spaces of ends") in your examples is exactly the discrete subset of $R^2$  (there is no difference between questions 0 and 1). More generally, if $A$ is a compact totally disconnected subset of $S^2$, then $S$ is homeomorphic to the space of ends of the surface $S^2-A$. The conclusion is that if you have two homeomorphic totally disconnected subsets $A, B\subset S^2$, then there is a homeomorphism of pairs $(S^2,A)\to (S^2,B)$. From this, you get an isotopy, since for homeomorphisms of closed surfaces homotopy is equivalent to isotopy. (In the case of the 2-sphere, this is due to Alexander, it is a corollary of  "the Alexander trick"). 
A: I don't know if this fits your idea of efficient, but problem 0 can be
solved (in $\mathbb{R}^n, n \ge 2$) by fairly simple 
geometry. 
Note that there is a $c \in \mathbb{R}^n$ such that no two
points of $S$ have the same distance to $c$. (The set of points for which
this is not the case form a meagre set of measure zero.) By compactness,
every closed ball about $c$ contains finitely many points of $S$, so the
points in $S$ can be enumerated by increasing distance from $c$.
This implies that there is a discrete family of concentric annuli, each
one containing a single point of $S$. Using spherical coordinates it is 
straightforward to construct a suitable isotopy for each of the annuli and
by the pasting lemma these can be combined into an isotopy of $\mathbb{R}^n$.
Alternatively, I suppose you could appeal to the path-connectedness of $SO(n)$ to get an interpolation. 
I would be surprised if this trick was
not known at least a century ago, but I don't know if anyone ever
bothered to write it down.
For problem 1 you obviously need the additional assumption that the unique
limit point is not in the set. Making that assumption, we can identify $\mathbb{S}^2$ with the one point compactification of $\mathbb{R}^2$ in
such a way that the limit point corresponds to $\infty$ and then the 
problem can be reduced to the previous one.
