I think you're confusing two different questions, which have related answers. I'll try to address them both.
Your primary question, judging from the comments, is this: Why must we represent an irrational number $x$ with a cut $(A,B)$ such that $A$ doesn't have a largest number? The answer comes down to what cuts are for. We use cuts to construct a set of numbers that extends the rationals. Every rational number must be represented in the new set of numbers; that's just part of what we're trying to do. Concretely, every rational number must correspond to exactly one cut.
Until we agree upon that goal, it's impossible to continue!
Now, there's a natural way to define the correspondence. For any rational number $q$, we can identify $q$ with either the cut $((-\infty,q],(q,\infty))$ or the cut $((-\infty,q),[q,\infty))$. Both cuts "point to" $q$.
This immediately has two consequences, which neatly answer your questions.
- If $(A,B)$ is a cut, and $A$ has a greatest element $q$, then the cut $(A,B)$ represents $q$. In particular, $(A,B)$ is a rational number. Turning this around, if $(A,B)$ is an irrational number then $A$ cannot have a greatest element!
- The cuts $((-\infty,q],(q,\infty))$ and $((-\infty,q),[q,\infty))$ are different sets, but they both correspond to the same rational number. In order to ensure that each rational number is represented by only one cut, we must discard one of the two! That is the real purpose of the third axiom. It states that $((-\infty,q],(q,\infty))$ is not a Dedekind cut. As a consequence, $q$ is represented by just one Dedekind cut, namely $((-\infty,q),[q,\infty))$.
Note that we could just as well disallow $((-\infty,q),[q,\infty))$ and accept $((-\infty,q],(q,\infty))$ as a Dedekind cut. Then the modified third axiom would state that $B$ must not have a smallest number. Which convention we choose is merely a matter of taste, so don't worry too much about the choice.