Here's a proof:
Let $ψ : G → \mathbb{C} ^{\times} $ be a group homomorphism. The following are equivalent:
$(1)$ $ψ$ is continuous;
$(2)$ the kernel of $ψ$ is open. If $ψ$ satisfies these conditions and $G$ is the union of its compact open subgroups, then the image of $ψ$ is contained in the unit circle $|z| = 1$ in $\mathbb{C}$.
Proof. Certainly (2) ⇒ (1). Conversely, let $N$ be an open neighbourhood of $1$ in $\mathbb{C}$. Thus $ψ^{−1}(N )$ is open and contains a compact open subgroup $K$ of $G$. However, if $N$ is chosen sufficiently small, it contains no non-trivial subgroup of $\mathbb{C} ^{\times} $ and so $K ⊂ Ker ψ$.
The unit circle $ S_1$ is the unique maximal compact subgroup of $\mathbb{C} ^{\times} $. If $K$ is a compact subgroup of $G$, then $ψ(K)$ is compact, and so it is contained in $S_1$. The final assertion follows.
I want to know why $K ⊂ Ker ψ$ then we can get (1)⇒ (2)?
Maybe a group contain a compact open subgroup will be a open group?
The proof is from the book《the local langlands conjecture for $GL(2)$ 》- colin j. bushnell, guy henniart,
Thank you for sharing your idea!