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.
