Consider a set $S$, and a total order $R$ over that set.
Part (a)
Given some element $e \in S$, explain why it is possible to partition $S$ into the following three sets: $$S_1 = \{ x \in S \mid (x,e) \in R \} $$ $$\{e\}$$ $$S_2 = \{ x \in S\mid (e,x) \in R \} $$
Why can you? I totally lost on this.
Part (b)
Given a partitioning of $S$ as stated in Part (a), explain why, for any $e_1 \in S_1$ and $e_2 \in S_2$, we know that both of the following are true: $$(e_1,e_2) \in R $$ $$ (e_2,e_1) \not\in R $$
Remember: You cannot use your intuitive understanding of ``less than,'' since the members of $S$ might not be numbers. Instead, use the formal definition of a total order, and argue that the statements above must be true!
Also how would not using inequalities work?
The partitioning follows directly from the definition of a total order (specifically, this is a strict total order, like < on the integers). Strict total orders exhibit trichotomy: any pair of objects is either in order one way, or the other, or else it is the same object. These three categories are exclusive, thus it partitions the set.
For the second part, assume that either case is not true, and apply transitivity of a total order at $e$.
EDIT: When I say "this is a strict total order" I am assuming that this is what you meant to say because it makes no sense otherwise. A (non-strict) total order does not imply the set can be partitioned, because it is reflexive (for all $a$, $a R a$ holds) therefore $e$ would be in all three sets and thus would not partition the set. (Reflexivity follows from antisymmetry and transitivity which are properties of a non-strict total order.)