I'm trying to understand the following theorem about the Pontryagin dual of a compact group $G$, and I'm completely lost on a single step.
Thm: Suppose that $G$ is compact. Then $\hat{G}$ is a discrete topological space.
Proof: Let $\varphi \in \hat{G}$ . Since $G$ is compact, we conclude that $\varphi(G) \subset \mathbb{T}$ is a compact subgroup of $\mathbb{T}$. Hence, if $\varphi$ satisfies: $$||\text{id}-\varphi||_{\infty} = \sup_{g \in G}|1-\varphi(g)| < \sqrt{3}$$ then we must have $\varphi(G) = \{1\}$, and thus $\varphi = \text{id}$. Thus, $\{\text{id}\}$ is an open neighborhood of $\text{id} \in \hat{G}$, proving the claim.
How does $\varphi$ satisfying that above inequality when $G$ is compact imply that $\varphi(G) = \{1\}$? Even if that was the case, how does this then imply $\{\text{id}\}$ is open? If $\varphi$ satisfies the inequality, then $\varphi \in B_{\sqrt{3}}(\text{id})$. I don't see how that implies anything they claim.
The point is that open balls are open by definition and $B_{\sqrt{3}}({\rm id}) = \{{\rm id}\}$.
In $\widehat{G}$, the open ball around the trivial character of radius $r$ is all $\varphi \in \widehat{G}$ such that $|\varphi(g) - 1| < r$ for all $g \in G$. So by definition of the topology on $\widehat{G}$, the set of all such $\varphi$ is an open subset of $\widehat{G}$.
The image $\varphi(G)$ is a subgroup of the unit circle, so if $|\varphi(g)-1| < r$ for all $g \in G$ then $\varphi(G)$ is a subgroup of the unit circle in which all of its elements satisfy $|z-1| < r$. Show the only subgroup of $\{z \in \mathbb T : |z-1| < \sqrt{3}\}$ is $\{1\}$, so if $r = \sqrt{3}$ then $\varphi(G) = 1$. Thus $\{\varphi : ||\varphi - 1|| < \sqrt{3}\} = \{{\rm id}\}$, so $\{{\rm id}\}$ is an open subset of $\widehat{G}$.
There is no important reason for using $r = \sqrt{3}$; that happens to be the biggest $r$ such that the only subgroup of $\{z \in \mathbb T : |z-1| < r\}$ is the trivial subgroup. We can't use larger $r$ because $\{z \in \mathbb T : |z-1| \leq \sqrt{3}\}$ contains the nontrivial group of cube roots of unity. It is probably more intuitive that for very small $r$ the arc $\{z \in \mathbb T : |z-1| < r\}$ has no nontrivial subgroup of the unit circle, and you could use that $r$ instead of going maximal by working with $r = \sqrt{3}$.