It holds that $W^{1,2}=H^1 \subset L^2 \subset H^{-1}$. This is clear since for every $v \in H^1(U)$, $u \mapsto (u,v)_{H^1}$ is an element of $H^{-1}$. Moreover for every $v \in L^2(U)$, $u \mapsto (u,v)_{L^2}$ is an element of $H^{-1}$. But I also know that $H^1$ is a Hilbert space and therefore it is isomorphic to its dual by the Riesz theorem.
My question is: how can there be $H^1(U) \subset L^2(U) \subset H^{-1}$ as well as $H^{-1}$ can be identified with $H^1(U)$?
You should not identify $H^{-1}$ with $H^1$, it leads to nothing but confusion. These spaces are dual to each other, but we do not think of the duality map as the identity map. The elements of these spaces have different meanings to us: $H^1$ consists of reasonably nice functions, $H^{-1}$ has some ugly distributions among its elements. So it makes sense that $H^1$ should be a proper subset of $H^{-1}$.
For that matter, all separable Hilbert spaces are isomorphic to each other, but this does not mean we should think that they are the same space.
But it's usually fine to identify $L^2$ with its dual. This allows for the following: the adjoint of the inclusion map $\iota:H^1\to L^2$ is a linear operator $\iota^*:(L^2)^* \to (H^1)^*$, that is, $\iota^*:L^2\to H^{-1}$. So we can interpret two inclusions $H^1 \subset L^2 \subset H^{-1}$ as adjoints of each other.