Existence of continuous path connecting points on a plane I've been thinking over this problem this weekend and although my investigations on it has led to other interesting theorems I am still nowhere close to solving it.

For any set of disjoint unordered pair of points (in $\mathbb{R}^2$), call it $S$, there is a set of simple continuous curves satisfying the following properties:
  
  
*
  
*The set of end points for each curve is a member of $S$ and vice-versa
  
*If two curves can intersect, then the intersection must be an end-point of one of the curves
  

I feel like I was able to give a proof for this when $S$ was countable using mathematical induction. Basically used that fact the plane remains connected when $|S|=1$ and if the plane is connected after adding a number of curves, it is possible to create an extra curve keeps it connected. So by induction if $S$ is countable then the constructed curves satisfy the given properties. I've never used mathematical induction in this manner so I'm not sure if this is a valid proof. 
Anyway, I can not think of a way to go about proving or giving a counterexample for an uncountable $S$. I'm stuck even when I let $S$ be the closed unit square (with interior) which I suspect fails the conjecture. What if $S$ is nowhere dense? Will that always satisfy the conjecture?
 A: By the Brouwer fixed-point theorem, "every continuous function $f$ from a convex compact subset $K$ of a Euclidean space to $K$ itself has a fixed point". Because $[0,1]$ is a compact, convex subspace of $\mathbb R$, your bijection is impossible.
Indeed, if you are generalising up to $K=[0,1]\times[0,1]$, it's still impossible.
A: Let $S=\{c_n=\{x_n,y_n\} \mid n \geq 1\}$ be a set of disjoint pairs of points in $\mathbb{R}^2$. For convenience, let $C_0= \bigcup\limits_{n \geq 1}c_n$.
Notice that $\mathbb{R}^2 \backslash C_0$ is path connected, because between any points $x,y \in \mathbb{R}^2 \backslash C_0$ there are uncountably many disjoint linear piecewise paths whereas $C_0$ is countable. Let $X_1$ be the range of a path between $x_0$ and $y_0$ in $\mathbb{R}^2 \backslash C_0$.
By induction, suppose we have built $X_n$ such that $X_n$ is the union of paths connecting the pairs $c_k$, for $1 \leq k \leq n$ as required. Again, $\mathbb{R}^2 \backslash (X_n \cup C_0)$ is path connected, so you can introduce $X_{n+1}$ as the union of $X_n$ and the range of a path connecting $x_{n+1}$ and $y_{n+1}$ in $\mathbb{R}^2 \backslash (X_n \cup C_0)$.
Finally, set $X= \bigcup\limits_{n \geq 1} X_n$. By construction, $X$ is a countable disjoint union of ranges of paths connecting the pairs $c_n$. Moreover, the union is increasing, so two curves don't intersect (otherwise, they would interset in some $X_n$, impossible by construction).
Added: For a counterexample when $S$ is uncountable, take $S= ...$ (under edition).
A: Proof of a counterexample when $S$ is uncountable. Using the complex plane, let $S=\{\{e^{i\theta}, e^{i(\theta+\pi)}\}~|~\theta \in [0,\pi) \}$. 
Suppose $\mathcal{F}$ is a set of simple curves that satisfy the hypothesis.


*

*By hypothesis, if two curves intersect, then the point of intersection is an endpoint for one of the curves.

*There are at most two curves which only intersect the circle at endpoints. All other curves must cross the circle at least once. This has been modelled in following image:

I'll assume this model but it can also be proven when there are only $0$ or $1$ which don't cross the circle. Without loss of generality, there are uncountably many curves $\mathcal{F}_1$ that join "red points" and cross $B_1$. Similarly, for these points of intersections, $I \subset B_1$, let $\mathcal{F}_2$ be set of uncountable curves that join them to $B_2$ and crossing $R_2$ (without loss of generality again).
$\mathcal{F}' = \mathcal{F}_1 \cup \mathcal{F}_2$.
By intermediate value theorem, the curves in $\mathcal{F}'$ cross the $x$-axis. For each point $p \in I$, there are two distinct point, $x_1$ and $x_2$, corresponding points on the $x$-axis (one for a curve in $\mathcal{F}_1$ and another for $\mathcal{F}_2$). Since the curves don't cross, there is no other intersection point between $x_1$ and $x_2$. Let $x_p$ be the midpoint of $x_1$ and $x_2$. This forms a bijection between $x_p$ and $p$. Each $x_p$ is isolated and any set of isolated points is countable.$^1$ This is a contradiction as there is also bijection from each function in $\mathcal{F}_1$ to a given $p$.
Therefore, no such $\mathcal{F}$ exists for the given $S$.
$^1$ One of the interesting theorems I proved in my earlier investigations. Glad I could finally use it somewhere.
