Antisymmetric Relations

Try this: consider a relation to be antisymmetric, UNLESS there exists a counterexample: unless there exists $(a, b) \in R$ and $(b, a) \in R$, AND $a\ne b$.

Since no such counterexample exists in for your relation, it is trivially true that the relation is antisymmetric.

Another way to put this is as follows: a relation is NOT antisymmetric IF AND ONLY IF there exist $a, b$ such that BOTH $\;(a, b)\in R\;$ AND $\;(b, a) \in R\;$ BUT $\;a\ne b$.

This is true of other properties as well: a property holds for a relation unless there exists a counterexample such that the property fails to hold. Put differently, a property FAILS to hold IF AND ONLY IF a counterexample exists.

$R$ is antisymmetric iff whenever both $(a,b)$ and $(b,a)$ are in $R$ then $a=b$.

In your example, there is no pair $(a,b) \in R$ that also has $(b,a) \in R$, so the statement is vacuously true.

Another (equivalent) way of looking at it is that $R$ is not antisymmetric iff there are elements $a,b$ with $a\neq b$ and both $(a,b),(b,a) \in R$.

In my opinion your misunderstanding is in the logic, not in the set theory.

If $p$ is false, then the conditional statement “if $p$ then $q$” is vacuously true. This does not mean that $q$ is true, it means the entire statement is true.

As @Doug Chatham says,

For example, consider the implication, "If 2+2=5, then you will pass the course." Since 2+2 is not 5, the statement is a true statement, regardless of whether or not you pass the course.

Regarding your question, if $(a,b)\in R$ and $(b,a)\in R$ together imply $a=b$, and either one (or both) of $(a,b)\in R$ or $(b,a)\in R$ is false, then the entire statement is true: R is antisymmetric.

The definition of antisymmetry does not state

if $(a,b) \in R$ then $(b,a) \in R$

(that would be the definition of symmetry).