$\newcommand{\lcm}{\operatorname{lcm}}$
If $s, t$ are solutions of $$\begin{cases} x \equiv a \pmod m, \\ x \equiv b \pmod n, \end{cases} $$ prove that $s \equiv t \pmod r$ where $r = \lcm(m, n)$
$s = a \pmod m$ and $s = b \pmod n$
$t = a \pmod m$ and $t = b \pmod n$
So,
$s - a = mk$ and $s - b = nq$
$t - a = mj$ and $t - b = nl$
Then,
$s - a - (t - a) = mk - nj = m(k - j)$
and $s - b - (t - b) = nq - nl = n(q - l)$
Thus,
$m, n \mid s - t$. Now let $s - t = q\lcm(m, n) + r$ with $0 \le r < \lcm(m, n)$. So, $r = (s - t) - q\lcm(m, n)$. Then $m, n \mid r$. If $r \neq 0$, then $r > \lcm(m, n)$ which is false. So, $r = 0$.
Does this proof work?