Is there a planar point set such that no matter how you colour the points with two colours can you can always find a triangle with exactly one point inside so that all four points have the same colour?
I'm not really sure how to start working on this graph problem any advice?
Yes, this can be done with a set of $22$ points. I apologize for the clumsy description below; I don't know how to draw pictures.
Start with a point $O$ and three points $P_1,Q_1,R_1$ spaced $120^\circ$ apart on a circle centered at $O$; e.g., $0$ and the three cube roots of unity. On the same circle choose points $P_0,P_2$ near $P_1$, points $Q_0,Q_2$ near $Q_1$, and points $R_0,R_2$ near $R_1$, so that each of the $27$ triangles $P_iQ_jR_k$ has $O$ in its interior. We may assume that the points are labelled $P_0,P_1,P_2,Q_0,Q_1,Q_2,R_0,R_1,R_2$ in counterclockwise order.
Call the two colours red and blue, and assume without loss of generality that $O$ is coloured red. If each of the three clusters contains a red point, then we have a red triangle $P_iQ_jR_k$ having the red point $O$ in its interior, and no other points; the additional $12$ points will be placed outside the circle. Thus we may assume that one of the clusters consists of three blue points. Without loss of generality, $P_0,P_1,P_2$ are coloured blue.
Next, choose points $U_0,U_1,U_2,U_3$ as follows: they are outside the circle, they all lie on a straight line parallel to the line segment $P_0P_2$, and the $12$ line segments $U_iP_j$ do not intersect the circle. Thus for each $i=0,1,2,3$, the triangle $U_iP_0P_2$ has the point $P_1$ in its interior. Now, perturb the points $U_i$ slightly, so that each $U_i$ still forms a triangle $ U_iP_0P_2$ with only the point $P_1$ in its interior, but instead of $U_0,U_1,U_2,U_3$ being collinear, one of them lies in the interior of the triangle formed by the other three. If $P_0,P_1,P_2$ are all blue then, whether one of the $U_i$ is blue or whether all four of them are red, we have the configuration we want.
Of course, we have to choose four more points $V_0,V_1,V_2,V_3$ which do the same trick if $Q_0,Q_1,Q_2$ are all of one colour, and four more points $W_0,W_1,W_2$ for $R_0,R_1,R_2$, making $22$ points in all.
Update: Well, actually, $13$ points are enough. Please consider the clumsy, inefficient $22$-point construction described above as a hint.