Worth emphasis: the common proof in Brian's answer is simply an ideal-theoretic form of the common Bezout-based proof of $ $ Euclid's Lemma $ $ for integers. To highlight the analogy we successively translate the Bezout-based proof into the language of gcds and (principal) ideals.

**Euclid's Lemma** in Bezout form, gcd form and ideal form

$\!{\!\begin{align}
\color{#0a0}{Ax\!+\!ay}=&\,\color{#c00}{\bf 1},\,\ A\ \mid\ ab\ \ \ \Rightarrow\, A\ \mid\ b.\ \ \ {\bf Proof}\!:\, A\ \mid\ Ab,ab\, \Rightarrow\, A\,\mid Abx\!\!+\!aby = \smash{(\!\overbrace{\color{#0a0}{Ax\!+\!ay}}^{\textstyle\color{#c00}{\bf 1}}\!)} b = b\\
\color{#0a0}{(A,\ \ \ a)}=&\,\color{#c00}{\bf 1},\,\ A\ \mid\ ab\ \ \ \Rightarrow\, A\ \mid\ b.\ \ \ {\bf Proof}\!:\, A\ \mid\ Ab,ab\, \Rightarrow\, A\,\mid (Ab,\ \ ab) = (\color{#0a0}{A,\ \ \ a})\ \ b =\, b\\
\color{#0a0}{A\!+\!(a)}=&\,\color{#c00}{\bf 1},\,\ A\supseteq\! (ab)\, \Rightarrow\, A \supseteq\! (b).\, {\bf Proof}\!:\, A \supseteq Ab,ab \,\Rightarrow A\supseteq Ab\!+\!(ab)\! =(\color{#0a0}{A\!+\!(a)})b =\! (b)\\
\color{#0a0}{A +{\cal A}}\ =&\,\color{#c00}{\bf 1},\,\ A\supseteq {\cal A B}\, \Rightarrow\, A \supseteq\, {\cal B}.\,\ {\bf Proof}\!:\, A\, \supseteq\! A{\cal B},\!{\cal AB}\!\Rightarrow\! A\supseteq A{\cal B}\!+\!\!{\cal AB} =(\color{#0a0}{A+{\cal A}}){\cal B} = {\cal B}
\end{align}}$

The third ideal form is precisely the same proof as in Brian's answer. The last form shows that the proof works more generally for coprime (i.e. comaximal) ideals $\,A,\, {\cal A},\, $ i.e. $\, A+{\cal A}= 1.\,$ In the second proof for integers, we can read $\,(A,a)\,$ either as a gcd or an ideal. Read as a gcd the proof employs the universal property of the gcd $\, d\mid m,n\iff d\mid (m,n)\,$ and the gcd distributive law $\,(Ab,ab) = (A,a)b.\,$ In the first proof (by Bezout) the gcd arithmetic is traded off for integer arithmetic, so the the gcd distributive law becomes the distributive law in the ring of integers.