Suppose for contradiction that the family is finite. Construct a graph where the discs are vertices and there is an edge between two vertices iff their corresponding discs are tangent. Evidently, is planar, so it satisfies Euler's characteristic , from which we may deduce . Hence the average degree of the vertices of is less than , which gives the required contradiction, since each disc is tangent to at least others.

We consider a graph with vertices representing the people. The edge is blue when people and know each other, and red otherwise. The problem is equivalent to showing that the edge-induced subgraph of blue edges in is -colorable. We start with a lemma.

Lemma. The edges of are colored red or blue in such a way that there is no blue and if is the edge-induced subgraph of blue edges, then . Then all the red edges of form a cycle of length .

Proof. We first claim that under the conditions of the lemma, there is no red . It is well known that (see here for a proof), so there is a monochromatic . Label the vertices of . Suppose, for contradiction, that there is a red , which, WLOG, we label . If some edge , where , is red, then the -partition , where , shows that , which is forbidden. Hence is a blue . .

Hence , otherwise we would have a blue . Suppose, WLOG, that are red. Then is a red , so that is a blue from . If now is blue, then is a blue . Otherwise, is red, and we have the -partition , so that , and we have our contradiction.

Therefore there is no red , and our claim is true.

We now show that all the red edges of form a cycle of length . WLOG, let be red. If all edges , where are red, then we have a -partition , which is forbidden. Hence, WLOG, let be blue. One of the edges in the vertex-induced subgraph must be red: WLOG, let this be . Then is blue (otherwise is a red ) and is blue (otherwise we have a -partition ).

If is red, then must be blue and is red. Then are all blue, so our red edges are , which form a cycle of length .

If is blue, then similar arguments produce a contradiction. Thus our proof of the lemma is complete.

Consider now the complete graph , with vertices . It is well known that (see here for a proof), and since there is no blue , there must be a red . WLOG, we let be the red . If the edge-induced subgraph of blue edges in the vertex-induced subgraph is -colorable, then we are done.

Otherwise, from the lemma, we can assume, WLOG, that are red. For , we have four cases depending on how many of the are blue.

Case 1. None of the are blue. Then we have the -partition .

Case 2. Exactly one of the is blue (WLOG ). Then at least three of the edges , for are red, otherwise we have a blue . WLOG, suppose is red when , so that we have a -partition .

Case 3. Exactly two of the are blue (WLOG ). Then as before, for at least three of the and three of the are red. We can easily find the desired -partition.

Case 4. All the are blue. By similar arguments to before, we can find a -partition.

Hence the edge-induced subgraph of blue edges in is -colorable, as required.