Is this correct:
If $a\equiv b \ \mathrm{mod}(n)$ and $m|n$, then $a\equiv b \ \mathrm{mod}(m)$.
Let $a=q_{1}n+r$, $b=q_{2}n+r$ and $n=mc$. Then we have \begin{align*} \frac{q_{1}mc+r -(q_{2}mc+r)}{mc}=\frac{q_{1}m+r -(q_{2}m+r)}{m}, \end{align*} which imply that $a\equiv b \ \mathrm{mod}(m)$.
Thank you in advance.
It is correct. One can establish the result more simply, or at least using fewer symbols. For note that $a\equiv b\pmod{n}$ means that $n$ divides $b-a$.
Since $m$ divides $n$, it follows that $m$ divides $b-a$, that is, $a\equiv b\pmod{m}$.
Remark: You handled the calculation perfectly correctly. However, in my experience, when one is beginning to do Number Theory, it is best to avoid calculations with "fractions." Writing $x=yz$ can be safer than writing $z=\frac{x}{y}$.
Yes we can also see the result by: $$a\equiv b \ \mathrm{mod}(n)\iff a=b+kn=b+k\alpha m=b+k' m\iff a\equiv b \ \mathrm{mod}(m)$$