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Here is a positive answer using connected sets, not polyhedra.

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 $\{ a_1, a_2, ..., a_k\}$ (wlog), we form the set $S_i$ as the union of $G_i$ with the $k$ segments

• out of $a_1$ corresponding to $2$-tuple $\{a_2, a_3, \ldots, a_k\}$
• $\vdots$
• out of $a_k$ corresponding to $2$-tuple $\{a_1, a_2, \ldots, a_{k-1}\}$

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

The same method can be used to embed any countable hypergraph in $\mathbb{R}^2$.

Here is a positive answer using connected sets whose boundary is not a Jordan curve.

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 $\{ a_1, a_2, ..., a_k\}$ (wlog), we form the set $S_i$ as the union of $G_i$ with the $k$ segments

• out of $a_1$ corresponding to $2$-tuple $\{a_2, a_3, \ldots, a_k\}$
• $\vdots$
• out of $a_k$ corresponding to $2$-tuple $\{a_1, a_2, \ldots, a_{k-1}\}$

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

The same method can be used to embed any countable hypergraph in $\mathbb{R}^2$.

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 $\{ a_1, a_2, ..., a_k\}$ (wlog), we form the set $S_i$ as the union of $G_i$ with the $k$ segments

• out of $a_1$ corresponding to $2$-tuple $\{a_2, a_3, \ldots, a_k\}$
• $\vdots$
• out of $a_k$ corresponding to $2$-tuple $\{a_1, a_2, \ldots, a_{k-1}\}$

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

The same method can be used to embed any countable hypergraph in $\mathbb{R}^2$.

Here is a positive answer using connected sets, not polyhedra.

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 $\{ a_1, a_2, ..., a_k\}$ (wlog), we form the set $S_i$ as the union of $G_i$ with the $k$ segments

• out of $a_1$ corresponding to $2$-tuple $\{a_2, a_3, \ldots, a_k\}$
• $\vdots$
• out of $a_k$ corresponding to $2$-tuple $\{a_1, a_2, \ldots, a_{k-1}\}$

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

The same method can be used to embed any countable hypergraph in $\mathbb{R}^2$.

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.

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 $\{ a_1, a_2, ..., a_k\}$ (wlog), we form the set $S_i$ as the union of $G_i$ with the $k$ segments

• out of $a_1$ corresponding to $2$-tuple $\{a_2, a_3, \ldots, a_k\}$
• $\vdots$
• out of $a_k$ corresponding to $2$-tuple $\{a_1, a_2, \ldots, a_{k-1}\}$

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

The same method can be used to embed any countable hypergraph in $\mathbb{R}^2$.