If $G = \langle a\rangle$ and $b$ $\in$ $G$, the order of $b$ is a factor of the order of $a$ Prof. Pinter's "A Book of Abstract Algebra" presents the following exercise from the "Cyclic Groups" chapter:


  If $G = \langle a\rangle$ is finite and $b$ $\in$ $G$, the order of $b$ is a factor 
  of the order of $a$


I believe that the proof is given by Theorem $2$ of this chapter:


  Every subgroup of a cyclic group is a cyclic.


Does this theorem provide an adequate proof to this exercise?
The theorem you have indicated argues that every subgroup is cyclic. The question requires something slightly different.

Assuming $G$ is a finite group, what can you say about the order of the subgroup $H=\langle b \rangle$? Try and use Lagrange's theorem after that point.
Here's an answer more in line with the concepts learned in Pinter's book up to that point (ie. no Lagrange's theorem etc). 

Let ord$(a) = n$. Therefore, there are $n$ elements in $G$. 

Then since $b \in G$, we have $b = a^i$ for some $1 \leq i \leq n$. This is because since $G = \langle a \rangle$ is cyclic, $G$ contains all the powers of $a$ and nothing else. 

So we have $b^n = (a^i)^n = (a^n)^i = e^i = e,$ 

showing that ord$(b) | n$. 

That is, ord$(b) |$ ord$(a)$.