If there are two different immediate successors of $x$ , $a\neq b$ , then $x<a$ , $x<b$ , and $(x,a)$ and $(x,b)$ are both empty, but since for the order either $a<b$ or $b<a$ , one of the sets $(x,a)$ or $(x,b)$ is non-empty, contradiction. Similarly for immediate predecessors. "At most one smallest or largest element"-part follows from a similar argument as well (for example, if there are two different smallest elements of a set, then one has to be smaller than the other).