I have seen other proofs for this but was hoping if someone could tell me if my proof looks okay, or if I am missing something:
$\overline{a}\ne 0\in \mathbb{Z}/n\mathbb{Z}$ is a zero-divisor $\iff$ there exists $\overline{r}\ne 0\in \mathbb{Z}/n\mathbb{Z}$ such that $\overline{ra} = \overline{0} \iff ra\equiv 0 \pmod n\iff ra = mn$ for some $m\in \mathbb{Z}\iff ra+mn = 2mn \iff gcd(a,n)\ne 1$. (since $2mn\ne 1$)
There are numerous errors. The 2nd and 3rd equivalents are missing existential quantifiers and should be as below, and the inferences after that are incorrect (it seems that you incorrectly apply Bezout's gcd identity, but without any justification of your claims it is impossible to know exactly what reasoning you intended). Below is a sketch of a correct proof starting as you did:
$$\begin{align} &\ a\,\text{ is a zero-divisor}\ \ \ \pmod{\!n}\\[.2em] \smash[t]{\overset{\rm def}\iff} &\,\exists\, r\!:\ ar\equiv 0,\ a,r\not\equiv 0\!\!\!\pmod{\!n}\\[.2em] \iff &\,\exists\, r\!:\ n\mid ar,\ \ \,n\nmid a,r\\[.2em] \smash[t]{\color{#0a0}{\overset{1\!\!}\Leftarrow}\!\color{#c00}{\underset{2}\Rightarrow}}\,\, &\ (a,n)> 1,\ \ \ \ n\nmid a \end{align}\qquad\qquad$$
$(\smash{\color{#0a0}{\overset{1\!\!}\Leftarrow}})\ $ Let $\,r = \frac{n}{(a,n)}.\,$ Then $n\mid ra = n(a/(a,n))$ and $\,n\nmid r,\,$ else $\,\frac{r}n = \frac{1}{(a,n)}\in\Bbb Z\Rightarrow (a,n)\!=\!1.\,$ $(\color{#c00}{{\underset{2}\Rightarrow}})\ $ By contrapositive: $\, (a,n)\!=\!1,\ n\mid ar\,\Rightarrow\, n\mid r\ $ by Euclid's Lemma.
Remark $ $ We can view the above as: $ $ if $\,f(x) := ax\equiv 0\pmod{\!n}$ has more roots $\,x\equiv r,0\,$ than its degree $(=1),\,$ then $\,n\,$ is composite, and we can find find a proper factor of $\,n\,$ via a quick gcd calculation, namely $\,(a,n).\,$ This generalizes to any polynomial $f(x)$ with integer coefficients that has more roots than its degree (over $\Bbb Z_n)$
