If we restrict our space to only rational numbers (no irrationals) then the open interval $(-\infty, 1)$ is an example of a dedekind cut.
1) It isn't empty and it is not all of $\mathbb Q$.
2) If $r < q$ and $q \in (-\infty, 1)$ then $r\in (\infty, 1)$
3) If $q \in (-\infty, 1)$ then there is an $r > q$ so that $r \in (\infty, 1)$.
To find such an $r$ in property 3): We know that $q < 1$ so $1 - q > 0$. And $1-q > \frac {1-q}2 > 0$. So let $r = q + \frac {1-q}2 < q + (1-q) = 1$. As $r$ is rational, $r \in (-\infty, 1)$ and $r >q$. So $q$ is not a maximal element.
No element in $(-\infty, 1)$ is maximal because for any rational number less than $1$, one can always find a larger rational number between it and $1$.
Reading the comments you seem be confusing "maximal" elements with "numbers that are as large or larger than all elements in the set". The difference being that a maximal element has to actually be an element in the set. In the comments you asked "Why is $1$ not the maximal element of $(-\infty, 1)$". The answer to that is because $1$ isn't IN $(-\infty, 1)$.
All $(-\infty, q)$ for rational $q$ are dedikind cuts. In fact the only difference between an open interval $(-\infty, q)$ and a dedekind cut is that a dedekind cut might not have a well defined endpoint. A dedekind cut will be something like $(-\infty, ????)$ but that $????$ might not be describable.
For example:. $D = \{q \in \mathbb Q|q < 0$ or if $q\ge 0$ then $q^2< 2\}$ is of the form $(-\infty, ???)$ but $???$ would have to be some sort of square root of $2$ and there's no such thing in the rational numbers so there is no way to write that.
Which gives a hint to the punchline of Dedekind's joke. Once we define the real numbers, the dedekind cuts are nothing more or less than $(-\infty, x)\cap \mathbb Q$ where $x$ is a real number. But have to define the dedekind cuts before we define the real numbers in order to define the real numbers in the first place.