Let $R$ be an integral domain. Show that a non zero, non unit element $c$ of $R$ is NOT reducible iff its only divisors are units of $R$ and elements of $R$ which are associates of $c$.
Let $a\in R$, $a\neq 0$ and $a$ not a unit.
I will proof your statement by using the following (quite common) definition of the irreducibility of $a$: Whenever $a = bc$ with $b,c\in R$, then at least one element of $b$ and $c$ is a unit in $R$.
Let $A$ be the statement "$a$ is irreducible" and $B$ the statement "Every divisor of $a$ in $R$ is either a unit or an associate of $a$".
We want to show $A\iff B$ or equivalenty, $$\neg A\iff \neg B.$$ We split the proof into the two implications.
,,$\neg B\implies\neg A$'': Assume there exists a divisor $c\in R$ of $a$ which is neither a unit nor associate to $a$. Since $c$ is a divisor of $a$, there a $b\in R$ with $a = bc$. Then $b$ is not a unit (otherwise, $a$ associated to $c$). So $a = bc$ where $b,c$ are no units. Thus, $a$ is reducible.
,,$\neg A\implies \neg B$'': If on the other hand $a$ is reducible, then $a = bc$ with $b,c$ elements of $R$ which are no units. Thus, $c$ is a divisor of $a$ which is neither a unit nor associate to $a$ (otherwise, $b$ is a unit).