It' easiest if you just put these into common language:
i) a cut isn't empty. a cut isn't the whole field Q.
ii) if a cut contains an element, it contains all the rationals less than that element as well.
This does not mean that every element is less than every other element. (Which wouldn't make any sense.) It just means once a cut contains an element then it will contain all those below as well.
iii) a cut doesn't have a largest element. (for any element in the cut you can find a larger in the cut.
Unstated conclusion: If q isn't in a cut then all r > q are also not in the cut.
CHEATING BY LOOKING AHEAD: A cut is simply the segment $(-\infty, r) $, restricted to the rationals, for any real r . That's all.
EXCEPT we can't SAY that because we haven't defined what a real number is yet. So we ABSOLUTELY can not say that. Although that is precisely what we are getting at.
A cut is all the rationals up to ... whenever the rationals stop being in a cut. For any cut then for any rational q, either q is in the cut, or q is too big for the cut.
Cuts basically "cut" the rationals into two parts, those small enough to be in the cut, and those too big to be in the cut.
All $(-\infty, q); q\in Q$ (restricted to Q) are cuts but these are aren't the only cuts. In $(-\infty, q)$ $q$ is the rational least upper bound of the cut. Some cuts don't have rational least upper bounds. The set $\{q| q \in Q, q^2 < 2\}$ is a cut without a least upper bound. It will be the cut, $(-\infty, \sqrt 2) \cap Q$ but we can't say that because $\sqrt 2$ is utterly undefined in Q.