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KConrad
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Since $\mathbf C^\times$ is a divisible group, Zorn’s lemma tells us for every abelian group $G$ and subgroup $H$ that each group homomorphism $H\to \mathbf C^\times$ extends (somehow) to a group homomorphism $G\to\mathbf C^\times$.

Use $G = F^\times$ and $H = \langle u\rangle$, where $u$ is in $\mathcal O_F^\times$ and not a root of unity, so $H$ is isomorphic to $\mathbf Z$. Start with the homomorphism $H\to \mathbf C^\times$ where $u^k \to 2^k$ and extend to a homomorphism with domain $F^\times$ by Zorn’s lemma. This homomorphism is not continuous since $\mathcal O_F^\times$ is compact but the homomorphism on this subgroup of $F^\times$ is discontinuous because its image in $\mathbf C^\times$ is unbounded and thus not compact.

KConrad
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