Just start from an arbitrary point as $P_1$ , and after determining $P_1 , ... , P_i$ for a $i<n$ , take $P_{i+1}$ in such a way that the length of $P_{i}P_{i+1}$ is the most lenght between $P_i$ and all other points distinct from $P_1 , ... , P_{i-1}$. And since if in a triangle $ABC$ , we have $\angle A \ge 90$ , then $BC$ has the most length between sides of the triangle , so all of those angles are acute.

Here's another solution: Consider the path with the minimized rotation through all hamiltonian paths on this $n$ points.(that is , the sum of angles $\angle P_{i-1}P_iP_{i+1}$ is minimum) We shall prove that this path satisfies the desired condition. Assume otherwise and consider somewhere , some angle $\angle BCD$ is non-acute where $ABCDE$ are 5 consecutive vertices in the path(the cases $n=3,4$ can be done by some caseworks easily). Now just consider this 3 paths : $ABCED , ABECD , ABDCE$ and using the fact that $CBD+BDC < BCD$ and the initial path minimized the sum of angles : $BED < 90 , BCE+BDE<CBD , BDC+FCD<DEC$ where $F$ is a point satisfying $FCB=FDB=CBD$, which exists uniquely , we'll get that $E$ should be inside the quadrilateral formed by $B,F,BD \cap CF$ and the infinity point on direction $BC$ , and also inside a circle with diameter $BD$ which can be easily seen that it wouldn't satisfy the third result we'd get from considering those paths. So we don't have bon-acute angle and we're done