What is missing in the original solution is the following observation: If $F([a,c])$ does not lie a hyperplane, then we can find $\epsilon>0$ such that the extremum of $F$ on any one-dimensional projection orthogonal to $F(c)-F(a)$ does not occur on $(c-\epsilon,c]$. Similarly for $[c,a]$.

With this, you just pick $n-2$ points on $(c-\epsilon,c+\epsilon)$. We may assume that the space spanned by $F'$ at each of these points is $n-1$-dimensional and does not contain the vector $F(c)-F(a)$ (if it necessarily does, then adding one additional point makes the collection linearly dependent). Now throw in $F(c)-F(a)$ to obtain a hyperplane $L$ and take an orthogonal projection to obtain an extreme point on each side of $c$. Its derivative is zero, and it doesn't lie in the original collection by the remark.

What is missing in the original solution is the following observation: If $F([a,t^\ast])$ does not lie a hyperplane, then we can find $\epsilon>0$ such that the extremum of $|P\circ F|$ on any one-dimensional projection $P$ orthogonal to $F(t^\ast)-F(a)$ does not occur on $(t^\ast-\epsilon,t^\ast]$. Similarly for $[t^\ast,a]$.

With this, you just pick $n-2$ points on $(t^\ast-\epsilon,t^\ast+\epsilon)$. We may assume that the space spanned by $F'$ at each of these points is $n-1$-dimensional and does not contain the vector $F(t^\ast)-F(a)$ (if it necessarily does, then adding one additional point makes the collection linearly dependent). Now throw in $F(t^\ast)-F(a)$ to obtain a hyperplane $L$ and take an orthogonal projection to obtain an extreme point on each side of $t^\ast$. Its derivative is zero, and it doesn't lie in the original collection by the remark.