Why are Berkovich spaces locally connected? A characteristic feature of Berkovich spaces is that they are locally connected (in fact, locally contractible). I'd like to understand the proof. The key ingredient seems to be Corollary 2.2.8 in Berkovich's book. A key step in the proof is to take a connected component $W_n$ of a certain set $V_n$, but there is no explanation of why this connected component should be open (as I think it needs to be). I'd like to understand why.
Question: Why are Berkovich spaces locally connected?
Ideally, I'd appreciate a clarification of Berkovich's proof, or else a pointer to some other proof in the literature. Also, I'd be happy to understand the case of the Berkovich spectrum of a ring.
 A: Let $X = \mathcal{M}(A)$ be an affinoid space. There is a map $\mathcal{M}(A) \to \operatorname{Spec}(A)$ sending a seminorm to its kernel. It is continuous, surjective and induces a bijection between the connected components of $\mathcal{M}(A)$ and $\operatorname{Spec}(A)$. Let me prove the last point.
First assume that $\operatorname{Spec}(A)$ is connected. Then $\mathcal{M}(A)$ is connected too. Indeed, consider a clopen subset $U$ of $\mathcal{M}(A)$. We can construct an analytic function $e$ on $\mathcal{M}(A)$ that is 0 on $U$ and 1 on the complement. But the global sections on $\mathcal{M}(A)$ are nothing but $A$ itself (Tate's theorem), so $e$ defines an element of $A$, which is idempotent. Since $\operatorname{Spec}(A)$ is connected, $A$ has only the trivial idempotents, so $e$ equals 0 or 1, hence $U$ is either the whole $\mathcal{M}(A)$ or the empty set.
For the general case, pick a connected component of $\operatorname{Spec}(A)$. It is a Zariski-closed subset of $\operatorname{Spec}(A)$, hence of the form $\operatorname{Spec}(B)$, with $B$ a quotient of $A$, so in particular an affinoid algebra. The previous argument shows that its preimage $\mathcal{M}(B)$ is connected. This finishes the proof of the bijection.
Finally, recall that the affinoid algebra $A$ is noetherian, so $\operatorname{Spec}(A)$ has finitely many connected components, and similarly for $\mathcal{A}$. As a consequence, those connected components are open.
I think this is what you need to complete the proof of local connectedness.
I am not sure what you mean about the Berkovich spectrum of a ring at the end of your post, so I will not comment on that.
