Let $R$ be an integral domain. If $x \in R$ is prime, then $x$ is irreducible. I am trying to understand the proof for the following theorem:

Let $R$ be an integral domain. If $x \in R$ is prime, then $x$ is irreducible.

Here is the proof:

I typed this a while ago and I don't understand the part where if $x | bc$, then $x=bc$? Is something wrong at this step?
Also, is the definition of prime elements where $p$ is prime if whenever $p|ab$, then either $p|a$ or $p|b$? Now I don't feel so sure. This could be the reason why I am not understanding the proof...

You have it reversed. What you seek is the inference $(1)\Rightarrow(2)$ below.
Theorem $\,\ (1)\,\Rightarrow\,(2)\!\iff\! (3)\ $ below, $ $ for a nonunit $p\neq 0$
$(1)\ \ \ \color{#c00}{p\ \mid\ ab}\  \Rightarrow\  p\:|\:a\ \ {\rm or}\ \ p\:|\:b\quad$ [Definition of $\:p\:$ is prime]
$(2)\ \ \ \color{#c00}{p=ab}\ \Rightarrow\ p\:|\:a\ \ {\rm or}\ \ p\:|\:b\quad$ [Definition of $\:p\:$ is irreducible, in associate form]
$(3)\ \ \ p=ab\ \Rightarrow\ a\:|\:1\ \ {\rm or}\ \ b\:|\:1\quad$ [Definition of $\:p\:$ is  irreducible, in $\rm\color{#0a0}{unit}$ form]
Proof $\ \ \ (1\Rightarrow 2)\,\ \ \ \color{#c00}{p = ab\, \Rightarrow\, p\mid ab}\,\stackrel{(1)}\Rightarrow\,p\mid a\:$ or $\:p\mid b.\ $ Hence prime $\Rightarrow$ irreducible.
$(2\!\!\iff\!\! 3)\ \ \ $ If $\:p = ab\:$ then $\:\dfrac{1}b = \dfrac{a}p\:$ so $\:p\:|\:a\iff b\:|\:1.\:$ Similarly $\:p\:|\:b\iff a\:|\:1.$
