The Pontryagin dual $\widehat{G}$ of a second countable locally compact Abelian group $G$ is second countable. There is the following characterisation:

Let $G$ be a locally compact Abelian group. Then $G$ is second countable if, and only if, the Pontryagin dual $\widehat{G}$ is second countable with respect to the compact-open topology on $\widehat{G}$.

I have difficulties to show that $\widehat{G}$ is second countable whenever $G$ is. The other direction is just an application of Pontryagin's duality theorem.
Any help or comment is highly appreciated.
 A: Let $\scr B$ be a countable basis of $G$. From your previous question, it follows that $\mathscr S = \{L(\overline V,U) \mid V \in \mathscr B \text{ open with compact closure },U \subset S^1 \text{ open}\}$ is a subbasis of $\widehat G$.
We will show that 
$$\mathscr S' = \{L(\overline V,U(a,b)) \mid V \text{ open with compact closure },a,b \in \Bbb Q,a<b\}$$
is also a subbasis of the dual group of $G$, where $U(a,b) = \{e^{i\theta} \mid a<\theta<b\}$. The basis generated by $\scr S'$ is countable, so we are done.

Clearly, $\mathscr S' \subset \mathscr S$, so that $\mathcal T_{\mathscr S'} \subset \mathcal T_{\mathscr S}$. To show the reverse inclusion, let $O \in \mathcal T_{\scr S}$. Write it as the union of some $B_a$'s where each $B_a$ is a finite intersection of $L(\overline V,U)$.
Then, it is sufficient to show that each $L(\overline V,U)$ belongs to $T_{\mathscr S'}$.
Pick any $\gamma \in L(\overline V,U)$.
Since $\gamma(\overline V)$ is a compact set contained in the open set $U = \bigcup_{i \in I} U(a_i,b_i) \subset S^1$, it is actually contained in a finite number of $ U(a_i,b_i) $:
$$\gamma(\overline V) \subset \bigcup_{k=1}^n U(a_{i_k},b_{i_k})$$
Then I let you show that $$\gamma \in \bigcap_{k=1}^n L(\overline V, U(a_{i_k},b_{i_k})) =: V_{\gamma} \subset L(\overline V,U).$$
Therefore $$L(\overline V,U) = \bigcup_{\gamma \in L(\overline V,U)} V_{\gamma} \;\in\; T_{\mathscr S'}.$$
