Let $E$ be a quadratic extension of a local nonarchimedean field $F$ of characteristic zero (and odd residual characteristic). Let $\sigma$ be a generator of the Galois group $G = Gal(E/F)$. I'm looking for an elementary proof that there are infinitely many distinct (unitary) characters $\chi$ of $E^\times$ such that ${}^\sigma \chi = \chi \circ \sigma \neq \chi$.

Here's a sketch: the character $\chi$ is Galois invariant, i.e., ${}^\sigma \chi = \chi$ if and only if $\chi$ is trivial on the kernel of the norm map $N_{E/F}: E^\times \rightarrow F^\times$. Let $K = \ker N_{E/F}$. The group $K$ is a closed subgroup of the compact group $U_E$ of units in $E^\times$. We can extend any non-trivial character $\tilde \chi$ of $K$ to $E^\times$ to obtain a nontrivial character $\chi$ of $E^\times$ such that ${}^\sigma \chi \neq \chi$.

The part of the argument that is missing is to show that either:

(a) the character group $\widehat K = Hom(K,S^1)$ of $K$ is infinite,

or, if (a) is false (?),

(b) if $\widehat K$ is finite, then we need to show that there are infinitely many distinct extensions of at least one $\tilde \chi \in \widehat K$ to $E^\times$

I expect that (a) is true. Any suggestions to address this fact (or a reference) would be greatly appreciated.