Let $x_1,x_2,...,x_7$ be the vertices of a heptagon, $H$. Clearly there must be a point inside the heptagon, call it $p_1$. Draw a line though $x_1,p_1$. One side will contain at least three vertices of $H$. These form a convex pentagon. Inside that is a new point $p_2$. Draw a line though $p_1,p_2$. On side of this line will also contain at least three vertices of $H$. They form a pentagon which must contain a third point $p_3$. Let $P_i$ be the half plane defined by the line though $p_j,p_k$ and not containing $p_i$, for all permutations $(i,j,k) = \sigma(1,2,3)$. It is clear that one of $P_1,P_2,P_3$ contains three vertices of $H$, say $P_1$. Then those three points along with $p_2,p_3$ form a new pentagon which must contain a new point $p_4$. So the number of points is not less that $7+4=11$. The construction isn't too difficult so I'll leave it to somebody else.