Let $D$ be an integral domain with a multiplicative norm $N:D \rightarrow \mathbb{N}$ such that
$$ p|b \implies N(p) \leq N(b) \ \ \ \forall p,b \in D. $$ with equality only if $b,p$ are associates.
Now I can show the following;
Let $p \in D$ be prime, that is
$$ \forall u,v \in D,\ \ \ \ p|uv \implies p|u \ \ \ \text{or}\ \ \ p|v $$
now we claim that $p$ is irreducible in $D$. We prove this by contradiction, so suppose that $p$ is irreducible. Then $p$ has proper divisors in $D$, that is,
$$ \exists u,v \in D:\ \ p =uv. $$ where $u,v$ are not units or associates of $p$.
Now this means that $N(p) = N(uv) = N(u)N(v)$ meaning that $N(p)\geq N(u)$ and $N(p)\geq N(v)$.
Now notice that $$ p = uv \implies p|uv $$
but now since $p$ is prime we need $p|u$ or $p|v$. Recall that $u,v$ are not units and in particular not associates of $p$ so we have
$$ N(p)< N(u) \ \ \ \text{or}\ \ \ N(p)<N(v) $$
which is a contradiction. So we must have that $p$ is irreducible in $D$
QUESTION
First of all, is this proof correct? If is not, why not? If it is correct, are there weaker conditions under which the implication
$$ p\ \text{prime} \implies p\ \text{ irreducible} $$
holds? Also if the proof above is incorrect what are some of the conditions that are also needed?
In an integral domain, it is already the case that every prime element is irreducible (the converse holds in a UFD).
Say $p$ is a prime element of an integral domain $A$. If $p=ab$ then $p|ab$ and hence $p|a$ or $p|b$. WLOG take $p|a$ then $a=pk$ for some $k\in A$, and hence $p=pkb$ and thus $k$ and $b$ are units. This is true for any such $a,b$ and hence $p$ is irreducible.
Your argument essentially contains this argument, but makes use of the unnecessary multiplicative norm.