For a group of order $110$, there must be a cyclic subgroup of order $2$, $5$ or $11$.
I am not sure if the statement is correct or not.
My approach:
For a given group we can have a subgroup $H = \langle a\rangle$ generated by element $a\in G\space(a \neq e)$
$H$ can have order $2,5 ,11,10,22,55,$ or $110$. (By langrange's theorem)
If $H$ has order $2, 5, \text{ or } 11$. we are done.
If $H$ has order $10$ then consider subgroup of $H$ generated by $a^5 = \langle a^5\rangle$. Which will definitely have order $2$. (because $a^{10} = e \implies (a^5)^2 = e$ and if $(a^5)^k=e$ for some $k<2$ then $order(a) \neq 10$)
Similar arguments can be given for case $22, 55 , \text{ and }110$.
Yes, your argument is correct. It's good that you checked both that $a^5$ had order dividing $2$ but also could not have smaller order.