I think that Andrej Bauer's answer doesn't really address the important point.
Obviously being way below is an order theoretic notion that does not depend on the identity
of the elements of the order but just on the order itself.  In particular, whenever you have any order with an element way below another, you can replace the smaller one by an infinite
set that fits into the order exactly as the original element.  Now there is an infinite set way below another set.  But this is artificial and doesn't really say anything.

Even in partial orders of sets an infinite set can be way below another set.
Identify each $r\in\mathbb R$ with the set $\{q\in\mathbb Q:q<r\}$, i.e., with the left half
of a Dedekind cut in $\mathbb Q$ that corresponds to $r$.

Now the order on $\mathbb R$ is just set-theoretic inclusion of these sets of rational numbers.  However, as Andrej Bauer points out, $0$ is way below $1$.
This does not conflict with James Cranch's answer, since the partial order of sets that we are looking at is not the full power set of $\mathbb Q$ but just a subordering of that, and in particular one that does not contain any finite set.