Let $[a, b] = m$ and $n$ be a common multiple of $a$ and $b$. Show that $m \mid n$.
We'll use Bézout’s Identity (1.9).
There are integers $x$ and $y$ such that $ax + by= \gcd (a, b)$ where $a$ and $b$ are not both zero.
We'll also use Proposition (2.22).
Let $a$ and $b$ be positive integers then $\gcd (a, b) × [a, b] = a × b$.
We're given $a \mid n$ and $b \mid n$. This means for some integers $x,y$ we have
$$\begin{align} ax &= n \\ \\ by &= n \end{align}$$
Bezout's identity gives us, for some integers $j,k$
$$ ja + kb = \gcd(a,b) $$
Using Proposition 2.2.2 to introduce the LCM
$$ [a,b] \times (ja + kb) = ab $$
Multiplying both sides by $n$ but using both $n=ax$ and $n=by$
$$ [a,b] \times (jaby + kbax) = nab $$
Dividing by $ab$
$$ [a,b] \times (jy + kx) = n $$
This tells us that $[a,b]=m$ is a factor of $n$. That is $m \mid n$.
Note: I found this online answer helpful (link).