We will work on the line segments intersecting the triangles, $\text {intersecting the surface of the tetrahedron=intersecting a face of tetrahedron}.$ For each triangle there $3$ other points, forming $3$ segments, therefore form a triangle. We have $2$ triangles, say $A $ and $B . $ Either $A $ pierces $B$ or the opposite $\implies $ there are $2$ segment or triangle intersections. We are not considering $\text {edge/edge} $ intersections if the points are in general position. Summing across all the pairs $A/B $ will give an even number. Observe that the sign of points in the plane of $A $ will be $2$ on one side and $1$ on the other side. They can each either be inside or outside the triangle for $2$ segment/plane intersections. Showing $\text {in/out,out/in,in/in} $ which are the chain link and piercing case.

Since no four points are coplanar there does not exist four points $A,B,C,D$ such that $AB$ intersect $CD$. Therefore, all the red points are distinct. Now for each two points, the line segment between them intersect the tetrahedron formed by the remaining points for exactly $0,1$ or $2$ times. We call a set of two points bad if there are exactly $1$ intersection and good otherwise. If there are no bad set then we are done, otherwise suppose some set $E,F$ is bad. Then we can WLOG assume $E$ lies inside tetrahedron $ABCD$ and $F$ lies outside it. Suppose $EF$ intersect the tetrahedron in the face $BCD$. Suppose $AF$, $AE$ intersect the plane $BCD$ at a point $G$ and $H$. Notice that $H$ lies in the interior of $\triangle BCD$, hence it lies inside exactly one of $\triangle BGD,\triangle BCG,\triangle CDG$, say $\triangle BGD$. We now claim that every set except $(E,C)$ is good which will complete the proof. Firstly, both $E,A$ lie on the different side of $F$ with respect to plane $BCD$, hence $(E,A)$ is good. Secondly, notice that $BECD$ lies inside $ABCD$, hence $A$ and $F$ both lies outside $BECD$, hence $A,F$ is good, by symmetry $(B,F)$, $(C,F)$, $(D,F)$ are good as well. Thirdly, Notice that both $A$ and $D$ lie outside $BCEF$, hence $(A,D)$ is good, similarly $(A,B),(A,C)$ are good. Fourthly, notice that $B$ and $C$ lie outside $AEFD$, hence $(B,C)$ are good, similarly $(C,D),(B,D)$ are good. Fifthly, notice that by assumption $E$, $B$ lie outside $AFCD$, hence $(B,E)$ is good, similarly $(E,D)$ is good. Sixthly, notice that $E\in ABGD\subset ABFD$ and $C$ is not in $ABFD$, hence $(E,C)$ is bad. This completes the proof.