A question about a proof in one of Sierpiński's papers The following is a question about Sierpiński's paper "Une démonstration du théorème sur la structure des ensembles de points", (link):
We call a set dense-in-itself if it does not contain any isolated points. An isolated point $x$ is a point for which a neighbourhood exists that does not contain points of the set other than $x$.
Let $C$ be a subset of $\mathbb R$ such that it does not contain any dense-in-itself subsets. Using an enumeration of the countable basis consisting of the balls $B(q,\frac{1}{n})$ we can show (without using the axiom of choice) that there exists a function $\varphi$ that picks a point $p$ in $C$. We can define $\varphi$ by observing that there is at least one ball that only contains one point of $C$. Define $\varphi$ to return the first point in the enumeration of the basic calls such that the ball contains no other points of $C$. 
$\varphi$ is a well-defined function on all subsets of $C$. 
It seems to me that $\varphi$ therefore is a choice function for $C$ and furthermore that it yields a enumeration of $C$. 
In the paper he sets $p_0 = \varphi (C)$ and then goes on to define families $\mathcal K$ of subsets of $C$ with the properties
(i)   $\{p_0\} \in \mathcal K$
(ii) if $A_i$ in $\mathcal K$ then $\bigcup_i A_i \in \mathcal K$
(iii) if $E \subsetneq C$ is in $\mathcal K$  then $E \cup \varphi (C \setminus E)$ is also in $\mathcal K$.
He uses $\mathcal K$ to give a proof that $C$ with the above property is enumerable without using the axiom of choice and without using transfinite induction. 

My question is: Does effective enumerability of $C$ not immediately
  follow from the fact that $\varphi$ is a choice function for $C$ (via
  transfinite induction)?

I suspect I might be using some form of choice in my thoughts without being aware of it. Thanks for your help. 
 A: My first attempt was completely incorrect.  Here is an outline of the proof, missing all of the details, and changing some notations.


*

*Demonstrate, given any scattered set $C$, how to effectively pick an element $\varphi (C) \in C$.

*Start with a scattered set $C$.

*Letting $p_0 = \varphi ( C )$, consider the smallest family $\mathcal{K}_0$ of subsets of $C$ satisfying the following:


*

*$p_0 \in E$ for all $E \in \mathcal{K}_0$;

*$\mathcal{K}_0$ is closed under arbitrary unions;

*Given any proper subset $E$ of $C$ in $\mathcal{K}_0$ the set $E \cup \{ \varphi ( C \setminus E ) \}$ belongs to $\mathcal{K}_0$.


*Show that $\mathcal{K}_0$ has the additional property that for any $E , G \in \mathcal{K}_0$ either $E \subseteq G$ or $G \subseteq E$ holds.

*Picking $p \neq p_0 \in C$ consider $E_p = \bigcup \{ E \in \mathcal{K}_0 : p \notin E \}$; note that $E_p \neq \emptyset$ and $E_p \in \mathcal{K}_0$.

*Note that $\varphi ( C \setminus E_p ) = p$ for all $p \neq p_0$ in $C$.

*Given distinct $p , p^{\prime} \in C \setminus \{ p_0 \}$ note that either $E_{p} \subseteq E_{p^{\prime}}$ or the reverse inclusion holds.  It then follows that either $p^{\prime} \in C \setminus E_{p}$ or $p \in C \setminus E_{p^{\prime}}$ (but not both).

*For each $p \in C \setminus \{ p_0 \}$ denote by $n ( p )$ the index of the first sphere $S_n$ for which $S_n \cap C \setminus E_p = \{ p \}$.

*Show that for distinct $p , p^\prime \in C \setminus \{ p_0 \}$ we have $n(p) \neq n(p^\prime)$.

*Enumerate $C \setminus \{ p_0 \}$ according to the values of $n(p)$.  (Add $p_0$ as a first element.)



Addendum:  The following would not have been possible without the comments made by Dave L. Renfro, below.  (Of course, any errors or mis-statements contained below are solely my fault.)
In his Cardinal and Ordinal Numbers, Sierpiński says the following:

If we can establish a 1-1 correspondence (at least one) between the elements of two given sets $A$ and $B$, then we say that the sets are effectively equivalent, and write $A \mathrel{\text{ef}\mathord{\sim}} B$.

What is meant here is that effective equivalence is a stronger notion than equivalence, where one must only demonstrate that two sets are in 1-1 correspondence without exhibiting any particular correspondence. (I guess this can be read somewhat intuitionistically, but also seems to parallel modern notions.)
More contemporaneous with the paper in question, in Les exemples effectifs et l'axiome du choix [Fund.Math., Tom.2 (1921), 112-118, link] Sierpiński gives the following definition.

Lorsque nous avons défini un objet particular $p$ jouissant de propriétés donnés $P$, nous disons que nous avons un exemple effectif d'un objet jouissant de propriétés $P$.  
[Google-Translate-aided translation: When we have defined a particular object $p$ enjoying the properties $P$, we say that we have an effective example of an object enjoying the properties $P$.]

So the effectiveness in the paper linked in the OP is only about exhibiting a particular 1-1 correspondence with the natural numbers.
A: I don't have access to Zermelo's paper at the moment, nor the ability to read French, but the argument you give here seems awfully similar to the one in Zermelo's 1908 paper in which he reproves the well-ordering theorem using $\Theta$-chains.
Despite by best effort I could not find when and where it was proved that $A$ can be well-ordered if and only if $\mathcal P(A)\setminus\{\varnothing\}$ has a choice function. I wouldn't be surprised if that was noted after Sierpinski wrote his paper, but I wouldn't be surprised if that was known before.
In either case, it is possible that Sierpinski is trying to show that $C$ is countable which is more than to say that it is well-orderable.
