An explicit construction for a "doubly weak" topology

Solution 1:

If $F$ is a family of linear functionals on a vector space $X$ then it induces a weak topology $w_F$ on $X$. The dual space of $(X,w_F)$ is the linear span of $F$ in the algebraic dual of $X$. In particular $(X,s)' = (X,w)'$ always holds.

This fully answers the first bullet. It also shows that there is no example as in the second bullet and that the procedure in the third bullet stabilizes at $(X,w)$ after one step.

The point is that if $\varphi \colon X \to \mathbb{F}$ is $w_F$-continuous then $U = \{x \in X \mid \lvert \varphi(x)\rvert \lt 1\}$ is a $w_F$-open neighborhood of $0$. By definition of the weak topology, there are $f_1,\dots,f_n \in F$ and $\varepsilon \gt 0$ such that $V = \bigcap_{i=1}^n \left\{x \in X \mid \lvert f_i(x)\rvert \lt \varepsilon\right\} \subseteq U$ and in particular $\bigcap_{i=1}^n \ker f_i \subseteq \ker \varphi $. Therefore $\varphi$ is a linear combination of the $f_i$ (a proof of this last assertion is here).

You can find a detailed proof in pretty much every book on topological vector spaces. A good reference is chapter 3 of Rudin's Functional Analysis, see in particular Lemma 3.9 and Theorem 3.10 (the Hausdorff assumption is only there because Rudin assumes locally convex spaces to be Hausdorff, as many authors do).