When you say that $H^s \subsetneq H^t$, what you're saying is that there is a distinguished compatible family of strict inclusions $f_{s,t} : H^s \subsetneq H^t$. Dualizing these inclusions gives maps $f_{s,t}^{\ast} : (H^t)^{\ast} \to (H^s)^{\ast}$.
These maps don't appear to be inclusions to me. At least, one can't conclude this from abstract properties of Banach spaces; judging by the finite-dimensional case I would guess that the dual of an inclusion tends to be a quotient. So, if you like, the argument fails here.
But I think there is a second point you're missing, so I'll press on. All Hilbert spaces are canonically anti-isomorphic to their duals $d : H \to H^{\ast}$ (anti-isomorphic means that scalar multiplication is sent to its conjugate), hence in particular there are anti-isomorphisms $d : H^s \to (H^s)^{\ast}$ and $H^t \to (H^t)^{\ast}$. (You can upgrade these to isomorphisms by composing with the canonical anti-isomorphism between a Hilbert space and its "opposite" Hilbert space, but I don't think you get anything from doing this, especially since all it tells you is that the "opposite" and dual are naturally isomorphic.)
There's no possible way to get a contradiction from an argument like this because the maps $f_{s,t}$ are completely different from the maps $d$. (This is what happens when mathematical notation obscures what's really important, which is the morphisms, not the objects.) Hilbert spaces which strictly include each other can still be abstractly isomorphic; remember that all separable Hilbert spaces are isomorphic to $\ell^2$.
Willie brings up a point in the comments that is an extra source of confusion here: the Sobolev spaces all have different inner products, but in the first part of the argument I guess you were implicitly thinking of all of them as subspaces of $L^2$ with the induced inner product? Unfortunately the notation here also hides the choice of inner product. It really is quite awful.