I'm trying to determine what a terminal object for the category Ord would look like. Right now I'm thinking that a terminal object in Ord is just a set containing a single element $a$ with the relation $a \leq a$. Denote this object by $O$.
Claim: If $A \in$ Ord, then we have a unique morphism $f : A \to O$ given by $f(a') = a$ for all $a' \in A$. I think this deduction means that every singleton set with an empty relation is also a terminal object in Ord.
If the claim is true how can I show it?
Given that the claim is true, I'm having trouble thinking of a global element $f : O \to K$ for some $K \in$ Ord, since $f$ must preserve the structure of $K$.
You are right about the terminal object. Why is there a unique morphism $A \to O$? A fortiori, there is a unique map $A \to O$ (since $O$ is a terminal object in $\mathbf{Set}$, if you want). It suffices to verify that this unique map is order-preserving, which is indeed the case: $x \leq y \implies a \leq a$ is a trivial implication.
There cannot be a nonempty initial object: if $O$ were a nonempty initial object, consider the disjoint union $K = O \sqcup O$: it admits at least $2$ order-preserving maps from $O$.
Instead, the initial object is the empty set with empty order-relation.