Solution:
Suppose four distinct points $P, Q, R, S$ (in that order on the circle) among these five are such that $\widehat{P Q}=\widehat{R S}$. Then $P Q R S$ is an isosceles trapezium, with $P S \parallel Q R$. We use this in our argument.

- If four of the five points chosen are adjacent, then we are through as observed earlier. (In this case four points $A, B, C, D$ are such that $\widehat{A B}=\widehat{B C}=\widehat{C D}$.) See Fig 1.

Fig 1.

Fig 2.

Fig 3.

- Suppose only three of the vertices are adjacent, say $A, B, C$ (see Fig 2.) Then the remaining two must be among $E, F, G, H$. If these two are adjacent vertices, we can pair them with $A, B$ or $B, C$ to get equal arcs. If they are not adjacent, then they must be either $E, G$ or $F, H$ or $E, H$. In the first two cases, we can pair them with $A, C$ to get equal arcs. In the last case, we observe that $\widehat{H A}=\widehat{C E}$ and $A H E C$ is an isosceles trapezium.

- Suppose only two among the five are adjacent, say $A, B$. Then the remaining three are among $D, E, F, G, H$. (See Fig 3.) If any two of these are adjacent, we can combine them with $A, B$ to get equal arcs. If no two among these three vertices are adjacent, then they must be $D, F, H$. In this case $\widehat{H A}=\widehat{B D}$ and $A H D B$ is an isosceles trapezium.

Finally, if we choose 5 among the 9 vertices of a regular nine-sided polygon, then some two must be adjacent. Thus any choice of 5 among 9 must fall into one of the above three possibilities.

Solution:
Here is another solution used by many students. Suppose you join the vertices of the nine-sided regular polygon. You get $\binom{9}{2}=36$ line segments. All these fall into 9 sets of parallel lines. Now using any 5 points, you get $\binom{5}{2}=10$ line segments. By pigeon-hole principle, two of these must be parallel. But, these parallel lines determine a trapezium.