I have to prove the following that every associate of an irreducible element is irreducible.
I am working in an integral domain so I will use the theorem which states that whenever $p$ is a non zero non unit element and when $p=rs$ then either $r$ or $s$ is a unit.
Proof:
Let $p$ be an irreducible element. with $p=rs$ then $r \ or \ s \ is \ a \ unit$
Now let the associate of the irreducible element $p$ be $a$ s.t. $a \in R$ then $a=pu$ where u is a unit. Hence $a=rsu$ with either r or s being units.
If $r$ is a nonunit then $a=s$ and then $s$ must be a unit from theorem.
If $s$ is a nonunit then $a=r$ where r must be a unit from theorem.
So from the theorem $a$ the associate must be irreducible?
I think you are better off by reformulating everything in terms of divisibility.
Namely, a non-zero, non-unit element $p$ is irreducible if whenever $r \mid p$, then either $r$ is a unit, or $r$ is associate to $p$.
Now recall that $q$ is associate to $p$ iff $p \mid q$ and $q \mid p$.
So let $p$ be irreducible, and consider $r \mid q$. Since $q \mid p$, you have $r \mid p$, so either $r$ is a unit, or it is associate to $p$, and thus to $q$.
If you want to argue your way, it goes like this. Suppose $a = r s$, so that $p = r (s u^{-1})$. Then either $r$ is a unit, or $s u^{-1}$ is a unit, and thus $s = (s u^{-1}) u$ is also a unit, as the product of two units.