Let $H$ be a vector space and $\{(\cdot, \cdot)_n\}$ a countable increasing set of inner products on $H$, i.e. $$(\phi, \phi)_1 \leq (\phi, \phi)_2 \leq \ldots, \quad \phi \in H$$ that are compatiable in the sense that if $\{\phi_i\} \in H$ and $\phi_i \rightarrow 0$ with respect to $\|\cdot \|_m$, then if $\{\phi_i\}$ is cauchy in $\|\cdot \|_n$ we also have that $\phi_i \rightarrow 0$ with respect to $\|\cdot \|_n$.
Define $H_n$ to be the completion of $H$ with respect to the norm $\|\cdot\|_n$ induced by the inner product $(\cdot, \cdot)_n$. Then $H$ is said to be countably Hilbert if it is complete relative to the topology generated by the neighborhood basis about $0 \in H$ given by: $$U_{n, \epsilon} = \{\phi \in H: \|\phi\|_n < \epsilon\}$$ for all $n \in \mathbb{N}$ and $\epsilon > 0$. It is easy to see that with this topology $H = \bigcap_{n=1}^\infty H_n$.
Now let $H_n'$ be the dual space of $H_n$. In volume 4 of Gelfand's book Generalized Functions he says:
...let $F$ be an element of the space $H_n'$. Then obviously $F$ is continuous relative to the topology on $H$, i.e. $F$ is an element of $H'$. From this it follows that $\bigcup_{n=1}^\infty H'_n \subset H'$. On the other hand, if $F$ is a linear functional on $H$, then as was mentioned in Section 1.2, $F$ is continuous relative to some of the norms, say $\|\phi\|_n = \sqrt{(\phi, \phi)_n}$, i.e. it belongs to the space $H'_n$. Thus $H' = \bigcup_{n=1}^\infty H'_n$. We note that the spaces $H_n'$ form an increasing chain $$H_1' \subset H_2' \subset \ldots.$$ In fact, since $(\phi, \phi)_m \leq (\phi, \phi)_n$ for $m \leq n$, then from the boundedness of a functional $F$ in the ball $(\phi, \phi)_m \leq 1 $ follows its boundedness in the ball $(\phi, \phi)_n \leq 1$, i.e. if $F$ belongs to $H_m'$ then it belongs to $H_n'$.
How does $F$ being continuous relative to the topology of $H$ imply that $\bigcup_{n=1}^\infty H'_n \subset H'$? How does Gelfand uses boundedness of $F$ to obtain the sequence of dual spaces above? Is there any intuitive way to see what's going on here?