I think that Andrej Bauer's answer doesn't really address the important point. Obviously being way below is an order theoretic notion, i.e., it is preserved under isomorphism, while being a finite set is not an order theoretic notion. 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.
However, even in partial orders of sets an infinite set can be way below another set. Identify each $r\in\mathbb R$ with the set $\{q\in\mathbb Q:q\le r\}$, 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 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.
It can be shown that every partial order is in fact isomorphic to a partial order of infinite sets ordered by inclusion, but the point in my example (or Andrej's example, for that matter) is that whenever $\mathbb P$ is a collection of sets such that $(\mathbb P,\subseteq)$ is isomorphic to $(\mathbb R,\leq)$, then the element corresponding to $0$ is way below the element corresponding to $1$, simply since this is an order theoretic notion, but $0$ is never represented by a finite set, since the corresponding element of $\mathbb P$ it has infinitely many sets below it, i.e., infinitely many subsets.