For an integer $a$, prove $[a]_n$ is a subset of $[a]_m$ if only if $n$ divides $m$ 
Let $a$ be an integer, prove $[a]_n$ is a subset of $[a]_m$ if only if $n|m$. 

I wanted to prove it by choosing an element $x$ and showing that if If $x≡ a \bmod m$ and $ x ≡ a \bmod n$, then $n$ divides $m$. 
So for the forward step can I assume $x=a+mk$ and $x=an+nr$, $k,r$   integer? And want to show $n=mq$. From assumption, I got $mk=nr$. How do I show $n=mq$, for some integer $q$?
 A: As has been pointed out in comments, from $x\equiv a\pmod{m}$ and $x\equiv a \pmod{n}$ we cannot conclude that $n$ divides $m$. 
It is not necessarily true that if $n$ divides $m$ then $[a]_n\subseteq [a]_m$. 
For example, let $n=2$, $m=4$, and $a=0$. Then $[a]_n$ consists of all multiples of $2$, and $[a]_m$ consists of all multiples of $4$. Certainly it is not the case 
that $[a]_n\subseteq [a]_m$. But it is true that $[a]_m\subseteq [a]_n$. When the proposed result is corrected, we can prove it using the intuition that you had.
Suppose first that $n$ divides $m$. Then if $x\equiv a\pmod{m}$, we can conclude that $x\equiv a \pmod{n}$. For if $m$ divides $x-a$ and $n$ divides $m$, then $n$ divides $x-a$. Thus $[a]_m\subseteq [a]_n$. (Note the direction of the containment.)
For the other direction, suppose that $[a]_m \subseteq [a]_n$. We show that $n$ divides $m$. 
So we are told that for every $x$ such that $m$ divides $x-a$, we have that $n$ divides $x-a$. Let $x=a+m$. Then $m$ divides $x-a=m$. We conclude that $n$ divides $x-a$, meaning that $n$ divides $m$.   
