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Let $\pi$ be a uniformizer of $F$, so $$ F = \pi^{\mathbf Z}(1+\mathfrak m_F) \cong \mathbf Z \times (1+\mathfrak m_F) $$ as topological groups, where $\mathbf Z$ has the discrete topology. Thus your question is the same as asking if all homomorphisms $1+\mathfrak m_F \to \mathbf C^\times$ are continuous.

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 $G = 1+\mathfrak m_F$ and $H = \langle 1+\pi\rangle$, which is isomorphic to $\mathbf Z$. Start with the homomorphism $H\to \mathbf C^\times$ where $(1+\pi)^k \to 2^k$ and extend to a homomorphism with domain $1+\mathfrak m_F$ by Zorn’s lemma. This homomorphism is not continuous since $1+\mathfrak m_F$ is compact but the homomorphism on this is discontinuous because its image in $\mathbf C^\times$ is unbounded and thus not compact.

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.

Let $\pi$ be a uniformizer of $F$, so $$ F = \pi^{\mathbf Z}(1+\mathfrak m_F) \cong \mathbf Z \times (1+\mathfrak m_F) $$ as topological groups, where $\mathbf Z$ has the discrete topology. Thus your question is the same as asking if all homomorphisms $1+\mathfrak m_F \to \mathbf C^\times$ are continuous.

Since $\mathbf C^\times$ is a divisible group, Zorn’s lemma tells us for an 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 $G = 1+\mathfrak m_F$ and $H = \langle 1+\pi\rangle$, which is isomorphic to $\mathbf Z$. Start with the homomorphism $H\to \mathbf C^\times$ where $(1+\pi)^k \to 2^k$ and extend to a homomorphism with domain $1+\mathfrak m_F$ by Zorn’s lemma. This homomorphism is not continuous since $1+\mathfrak m_F$ is compact but the homomorphism on this is discontinuous because its image in $\mathbf C^\times$ is unbounded and thus not compact.

Let $\pi$ be a uniformizer of $F$, so $$ F = \pi^{\mathbf Z}(1+\mathfrak m_F) \cong \mathbf Z \times (1+\mathfrak m_F) $$ as topological groups, where $\mathbf Z$ has the discrete topology. Thus your question is the same as asking if all homomorphisms $1+\mathfrak m_F \to \mathbf C^\times$ are continuous.

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 $G = 1+\mathfrak m_F$ and $H = \langle 1+\pi\rangle$, which is isomorphic to $\mathbf Z$. Start with the homomorphism $H\to \mathbf C^\times$ where $(1+\pi)^k \to 2^k$ and extend to a homomorphism with domain $1+\mathfrak m_F$ by Zorn’s lemma. This homomorphism is not continuous since $1+\mathfrak m_F$ is compact but the homomorphism on this is discontinuous because its image in $\mathbf C^\times$ is unbounded and thus not compact.

Let $\pi$ be a uniformizer of $F$, so $$ F = \pi^{\mathbf Z}(1+\mathfrak m_F) \cong \mathbf Z \times (1+\mathfrak m_F) $$ as topological groups, where $\mathbf Z$ has the discrete topology. Thus your question is the same as asking if all homomorphisms $1+\mathfrak m_F \to \mathbf C^\times$ are continuous.

Now take $F = \mathbf Q_p$. Using the $p$-adic exponential and logarithm, $1+p\mathbf Z_p \cong p\mathbf Z_p$ as topological groups when $p\not=2$ and $$ 1+2\mathbf Z_2 \cong \{\pm 1\} \times (1+4\mathbf Z_2) \cong \{\pm 1\} \times 4\mathbf Z_2. $$ Since $p\mathbf Z_p \cong \mathbf Z_p$ and $4\mathbf Z_2 \cong \mathbf Z_2$ as topological groups, your question when $F = \mathbf Q_p$ for any prime $p$ is equivalent to asking if every group homomorphism $$ \mathbf Z_p \to \mathbf C^\times $$ is continuous. Do you agree that there are a lot of discontinuous homomorphisms of that kind?

Explicitly, because $\mathbf C^\times$ is a divisible group, Zorn’s lemma tells us each group homomorphism $\mathbf Z\to \mathbf C^\times$ extends (somehow) to a group homomorphism $\mathbf Z_p\to\mathbf C^\times$. So start with the mapping $\mathbf Z\to \mathbf C^\times$ where $n\mapsto 2^n$ and extend it to a homomorphism on $\mathbf Z_p$. That extension is not continuous since its image is unbounded and thus not compact.

Let $\pi$ be a uniformizer of $F$, so $$ F = \pi^{\mathbf Z}(1+\mathfrak m_F) \cong \mathbf Z \times (1+\mathfrak m_F) $$ as topological groups, where $\mathbf Z$ has the discrete topology. Thus your question is the same as asking if all homomorphisms $1+\mathfrak m_F \to \mathbf C^\times$ are continuous.

Since $\mathbf C^\times$ is a divisible group, Zorn’s lemma tells us for an 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 $G = 1+\mathfrak m_F$ and $H = \langle 1+\pi\rangle$, which is isomorphic to $\mathbf Z$. Start with the homomorphism $H\to \mathbf C^\times$ where $(1+\pi)^k \to 2^k$ and extend to a homomorphism with domain $1+\mathfrak m_F$ by Zorn’s lemma. This homomorphism is not continuous since $1+\mathfrak m_F$ is compact but the homomorphism on this is discontinuous because its image in $\mathbf C^\times$ is unbounded and thus not compact.