I'm a bit confused. Consider the relation $<$ on $\mathbb N$. Then $<$ is per definition antisymmetric if
$$a<b \land b<a \implies a=b.$$
The problem which I have is that the premise of the implication can never be true. It is not possible for a natural number to be less and greater than another number at the same time. But when the premise is false (which is the case), the implication is always true. So it follows that $<$ is an antisymmetric relation - which is obviously wrong. Where is the fallacy?
It is antisymmetric. The statement
is true. It is true because, for every $x$ and $y$, the premise is false, but where's the problem?
Look at it from a different direction. Can you prove it is not antisymmetric? You should find $x$ and $y$ such that $x<y$, $y<x$ and $x\ne y$.
Are you perhaps surprised in learning that equality is antisymmetric? I hope not. Indeed, equality is a partial order relation, as well as an equivalence relation. Just apply the definitions.
Can you? Of course not. So it is antisymmetric.