Let $k$ be a local field, and consider the map $\phi: k^+ \hookrightarrow \widehat {k^+}$ given by $\eta \mapsto \eta X(\cdot)=X(\eta \cdot)$ where $X$ is a non-trivial character.
Tate argues in his thesis that $\phi$ has dense image simply because we have the implication $X(ηξ)=1 ∀η⟹ξ=0$. However, I am not sure how this shows that we can multiply $X$ by some $\eta$ to get arbitrarily close of another given $X'$. For that we would have to show that for any finite collection of compact subsets $B_{1,...n}$ in $k^+$ and open sets $U_{1...n}\subset S^1$ such that $X'(B_i)\subset U_i$ we can find $\eta$ such that $X(\eta B_i) \subset U_i$ as well.
If $C$ is the closure of the image of $\phi$, then $\widehat{k^+}/C$ is a locally compact Hausdorff abelian group under the quotient topology. To show $C$ is all of $\widehat{k^+}$, we just need to show that every character on $\widehat{k^+}/C$ is trivial. Let $\psi: \widehat{k^+}/C \to S^1$ be a character. Then we get a character on $\widehat{k^+}$ by composing:
$$\widehat{k^+} \stackrel{quot}{\to} \widehat{k^+}/C \stackrel{\psi}{\to} S^1.$$
By the Pontryagin duality isomorphism $k^+ \cong \widehat{\widehat{k^+}}$, this character is of the form $Y \mapsto Y(\zeta)$ for some $\zeta \in k^+$. Moreover, since this character factors through the quotient, it vanishes on $C$, which is equivalent to its vanishing on the image of $\phi$, which is precisely to say that $X(\eta\zeta) = 1$ for every $\eta \in k^+$. If we suppose Tate's implication holds, we infer that $\zeta = 0$, so that the character $\psi$ had to be trivial, as desired.