A paradox on Hilbert spaces and their duals

A key question here is "What precisely does $H_2\subseteq H_1$ mean"? Let's begin with the identity map on the pre-Hilbert space $H$: $$\text{id}:(H,(\cdot,\cdot)_2)\to (H,(\cdot,\cdot)_1).$$ where we assume that, for all $h\in H$, we have $(h,h)_1\leq (h,h)_2$. This map lifts uniquely to the completions $$i: H_2\to H_1.$$ It is often overlooked, however, that the resulting map $i$ may not be injective. The map $i$ is injective if and only if the bilinear form $((h,h)_2, H)$ is closable on $H_1$.

Let's suppose that $i$ is injective so that we can really think of elements of $H_2$ as also belonging to $H_1$. We should consider $H_2\subseteq H_1$ as shorthand for $i:H_2\to H_1$.

Now $i$ is a linear map, and the dual map gives $i^\prime:H_1^\prime\to H_2^\prime$. This gives a precise meaning to $H_1^\prime\subseteq H_2^\prime$. Combining this with Riesz representations on $H_1$ and $H_2$, we can write $$H_1\overset{j}{\rightarrow} H_1^\prime \overset{i^\prime}{\rightarrow} H_2^\prime \overset{k}{\rightarrow} H_2.$$ I think that this is what you mean by $H_1\subseteq H_2$.

In short, all of the paradoxes disappear when you keep careful track of the mappings involved.