This is exercise 2.98 in Artin's algebra textbook.
Use Proposition (2.6) to prove the Chinese Remainder Theorem: Let $m,n,a,b$ be integers, and assume the greatest common divisor of $m$ and $n$ is $1$. Then there is an integer $x$ such that $x \equiv a \text{ (modulo $m$)}$ and $x \equiv b \text{ (modulo $n$)}$.
Proposition 2.6 states the following:
Let $a,b$ be integers, not both zero, and let $d$ be the positive integer which generates the subgroup $a\mathbb{Z} + b\mathbb{Z}$. Then
(a) $d$ can be written in the form $d = ar + bs$ for some integers $r$ and $s$.
(b) $d$ divides $a$ and $b$.
(c). If an integer $d$ divides $a$ and $b$, it also divides $d$.
Below is my attempt. I'm hoping someone can check this over.
By Proposition (2.6), since $(m,n) = 1$, there exist integers $s,t \in \mathbb{Z}$ such that $sm + tn = 1$. Multiplying by $b-a$, we get \begin{align*} s(b-a)m + t(b-a)n = b - a. \end{align*} Let $y = s(b-a)m + a$. So we have $y + t(b-a)n = b$. Then $y - b = t(b-a)n$, so $y \equiv b \text{ (mod $n$)}$. Furthermore, subtracting $a$ from both sides and rearranging, we have \begin{align*} y - a = -t(b-a)n + (b - a) = -t(b-a)n + s(b-a)m + t(b-a)n = s(b-a)m, \end{align*} so $y \equiv a \text{ (mod $m$)}$, giving a solution to this system of congruences.