On the properties of the Sobolev Spaces $H^s$ Let $H^s(\mathbb{R}^d):= \{ u \in \mathcal{S}' : (1+|\xi|^2)^{s/2}\hat{u}(\xi) \in L^2(\mathbb{R}^d)\}$. It can be shown that this space is a Hilbert space and that $H^s \subset H^t$ if $t \leq s$. 
Now suppose we have $t > s>0$ such that $H^s \subsetneq H^t$. Then we know that $(H^s)^* \supsetneq (H^t)^*$, where $^*$ denotes the dual. 
At the same time, since $H^s$ is a Hilbert space, the Riesz Representation theorem tells us that $H^s \cong (H^s)^*$ and similarly, $H^t \cong (H^t)^*$. 
But then doesn't $(H^s)^* \supsetneq (H^t)^*$ imply that $H^s \supsetneq H^t$, which contradicts our earlier statement that $H^s \subsetneq H^t$?
What part of the argument here falls apart? 
 A: 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. 
A: All the Sobolev spaces $H^s$ are isomorphic as Hilbert spaces simply because they have the same dimension.. the isomorphism from $H^s$ to $H^t$ is given by multiplying the Fourier transform by $(1 + |\xi|^2)^{s - t \over 2}$. 
When one says that $H^s \subset H^t$ for $s > t$ one is looking at the finiteness of $||f||_{H_s} = ||(1 + |\xi|^2)^{s \over 2}\hat{f}(\xi)||_{L^2}$ and similarly for $H_t$. If you want a natural dual spaces that reverse the inclusion you have to view them as subspaces of a fixed Hilbert space with a single inner product, say that of $H^t$. In this case one can view the dual space of $H^s$ as $H^{-s}$ and the dual space of $H^t$ as $H^{-t}$ by taking $<f, g> = \int \hat{f}(\xi)\bar{\hat{g}(\xi)}\,d\xi$. In this situation the reversal of inclusions is clear. 
