Connected is probably too weak for what you have in mind.  But here's a fun solution for $k$-tuples of $n$ points in the plane.

We want $m={n \choose k}$ connected sets, $S_1, S_2, \ldots , S_m\subseteq \mathbb{R}^2$.  I'll want to associate each of these sets with a $k$-tuple, so I'll just refer to the $i^\mathrm{th}$ $k$-tuple.

Start with the graphs $G_i$ of $f_i (x) = \sin({1\over x})+ {i\over 2m}$ for $x > 0$ and $i = 1, 2, \ldots, m$.  Each of these graphs has 
the entire interval $I = \{ 0\} \times [{1\over 2}, 1]$ in its set of limit points.  Break $I$ into $n$ subintervals $I_1, I_2, \ldots, I_n$ (ordered
with increasing $y$-values.

Next, line up your $n$ points $a_1, a_2, \ldots, a_n$ along the line $x=-1$ (ordered with increasing $y$-values) and from $a_j$ draw ${n-1\choose 2}$ distinct line segments to the interval $I_j$.

Now for a the $i^\mathrm{th}$ $k$-tuple $\{ a_1, a_2, a_3\}$ (wlog), we form the set $S_i$ as the union of $G_i$ with the three segments

 - out of $a_1$ corresponding to $2$-tuple $\{a_2, a_3\}$
 - out of $a_2$ corresponding to $2$-tuple $\{a_1, a_3\}$
 - out of $a_3$ corresponding to $2$-tuple $\{a_1, a_2\}$

The fact that these sets are connected is the standard use of the topologist's sine curve.  They plainly do not intersect.