Unique way to topologise finite algebra over Huber ring Let me start with the following Lemma.

$\textbf{Lemma}$ Let $A$ be a Tate ring, and let $f\colon A\to B$ be a finite $A$-algebra. Then there is a unique way to topologise $B$ turning it into a Huber ring, making the structure morphism $A\to B$ continuous and such that there exists rings of definition $A_0$, $B_0$ of $A$ and $B$, respectively, with $f(A_0)\subset B_0$ such that the induced map $A_0\to B_0$ is finite.
$\textbf{Proof.}$ $\textit{Existence}$: Give $B$ the canonical $A$-module topology, i.e. give it the quotient topology for any surjection $A^n\twoheadrightarrow B$ of $A$-modules (this does not depend on the choice). Let $A_0$ be a ring of definition and let $\pi\in A_0\cap A^\times$ be a pseudo-uniformizer. Choose a finite number of generators $\{b_i\}$ of $B$ as an $A$-module s.t. $1$ is one of them, with each $b_i$ being integral over $A_0$. Denote by $M$ the $A_0$-submodule of $B$ generated by $\{b_i\}$. Then the integrality allows to choose $m\gg 0$ such that $B_0:=A_0[\pi^mb_1,\ldots,\pi^m b_n]\subset M$. Since also $\pi^m M\subset B_0$, the subspace topology on $B_0$ is the $\pi$-adic topology. Thus, $B$ is a Huber ring with ring of definition $B_0$. Clearly the structure morphism is continuous and the induced map $A_0\to B_0$ is finite as $B_0$ is integral and of finite type over $A_0$.  
$\textit{Uniqueness}$: Since $B_0$ is finite over $A_0$, there exists $m\gg 0$ such that $\pi^m B_0\subset M$. On the other hand, since the structure morphism is continuous, $\pi$ is topologically nilpotent in $B$ and so $\pi^N M\subset B_0$ for $N\gg 0$. This shows that the topology on $B$ is the one uniquely determined by making $M$ with the $\pi$-adic topology an open subgroup, which is precisely the canonical topology.



*

*I am wondering if the condition on the existence of $A_0,B_0$ making $A_0\to B_0$ finite is automatically satisfied knowing that the structure morphism is continuous and finite étale. Namely, in Remark 7.2 of these notes by Matthew Morrow, the uniqueness in case $A\to B$ being finite étale is stated without any finiteness condition on the level of rings of definition.  

*In Huber's book "Étale Cohomology of Rigid Analytic Varieties and Adic Spaces" it says in $(1.4.2)$ that for a (complete) Huber ring (it does not say Tate) and $A\to B$ a finite algebra, the canonical $A$-module topology on $B$ turns it into a Huber ring. My proof above clearly makes use of the existence of a topologically nilpotent unit, and I would like to know what the general argument looks like. Huber refers to $(3.12.10)$ of his "Bewertungsspektrum und rigide Geometrie", which is unfortunately not available to me.
 A: I think I can prove statement 2. (I too do not have access to this book "Bewertungsspektrum und rigide Geometrie" and was trying to look it up...)
Let $B$ be generated as an $A$-module by $b_1,\dots,b_n$.
Write $b_ib_j=\sum a_{ijk}b_k$ for $a_{ijk}\in A$.
Pick a small enough ideal of definition $I$ of $A_0$ such that $Ia_{ijk}\subset A_0$ for all $i,j,k$. Then we see that
$$
B_0=A_0+Ib_1+\dots+Ib_n
$$
is a subring of $B$.
Now note that if $U\subset A^n$ is open additive subgroup containing $0$, then $q^{-1}(q(U))$ is an additive subgroup of $A^n$ containing $U$, so it is open, and thus $q(U)$ is open. We now see that $B_0$ is an additive subgroup of $B$ containing the open additive subgroup $q(I\times\dots \times I)$, so $B_0$ is open.
Now if $V\subset B$ were an open neighborhood of $0$, then $q^{-1}(V)$ contains $I^N\times \dots \times I^N$ for some large $N$, so $U$ contains $q(q^{-1}(V))=I^Nb_1+\dots+I^Nb_n$. Write $1=a_1b_1+\dots+a_nb_n$ for $a_i\in A$. Pick a large enough $M$ such that $M>N-1$ and $I^Ma_i\subset I^N$. Then
$$
I^MB_0 = I^MA_0+I^{M+1}b_1+\dots+I^{M+1}b_n \subset I^Nb_1+\dots+I^Nb_n\subset V.
$$
We conclude that the ideals $(IB_0)^m$ form an open neighborhood basis of $0$ in $B$. Thus $B$ is a Huber ring with ring of definition $B_0$ and ideal of definition $IB_0$.
