Assume $Y$ is adic*. Then $X$ is adic* by Lemma \ref{lemma-property-goes-up-affine}. \medskip\noindent Assume $X$ is adic*. Write $X = \text{Spf}(A)$ for some adic ring $A$ which has a finitely generated ideal $I$ such that $\{I^n\}$ is a fundamental system of open ideals. By Lemmas \ref{lemma-iff-countably-indexed} we see that $Y$ is countably indexed. Thus, by Lemma \ref{lemma-countably-indexed}, we can write $Y = \text{Spf}(B)$ where $B$ is a weakly admissible topological ring with a countable fundamental system $\{J_m\}$ of weak ideals of definition. By Lemma \ref{lemma-morphism-between-formal-spectra} the morphism $X \to Y$ corresponds to a continuous ring map $B \to A$ which is taut by Lemma \ref{lemma-representable-affine}. Our first goal is to reduce to the case where $J_m A$ is equal to $I^m$. \medskip\noindent Set $Y_m = \Spec(B/J_m)$ so that $Y = \colim Y_m$. The scheme $Y_m \times_Y X$ is affine (Lemma \ref{lemma-affine-representable-by-algebraic-spaces}) and we have $X = \colim Y_m \times_Y X$. Say $Y_m \times_Y X = \Spec(A_m)$ so that $B/J_m \to A_m$ is a faithfully flat ring map. It follows from Lemma \ref{lemma-fibre-product-affines-over-separated} that $\Ker(A \to A_m)$ is the closure of $J_mA$. \medskip\noindent Choose $n \geq 1$. There exists an $m$ such that $\Spec(A/I^n) \to Y$ factors through $Y_m$. In terms of ideals \begin{equation} \label{equation-first} \forall n\ \exists m,\ J_m A \subset I^n. \end{equation} Choose $m \geq 1$. We can find an $n$ such that the morphism $\Spec(A_m) \to X$ factors through $\Spec(A/I^n)$. In terms of ideals \begin{equation} \label{equation-second} \forall m\ \exists n,\ I^n \subset \Ker(A \to A_m). \end{equation} Given an $m$ we can pick an $n = n(m)$ such that $I^n \subset \Ker(A \to A_m)$ by (\ref{equation-second}). Choose generators $f_1, \ldots, f_r$ of $I$. For any $E = (e_1, \ldots, e_r)$ with $|E| = \sum e_i = n$ write $$ f_1^{e_1} \ldots f_r^{e_r} = \sum g_{E, j} a_{E, j} + \delta_E $$ with $g_{E, j} \in J_m$, $a_{E, j} \in A$, and $\delta_E \in I^{n + 1}$ (possible by the above). Let $J = (g_{E, j}) \subset J_m \subset B$. Then we see that $$ I^n \subset J A + I^{n + 1} $$ As $I$ is contained in the Jacobson radical of $A$ and $I^n$ is finitely generated we see that $I^n \subset JA$ by Nakayama's lemma. More precisely apply part (2) of Algebra, Lemma \ref{algebra-lemma-NAK} to see that $M = (I^n + JA)/JA$ is zero. \medskip\noindent We first apply what we just proved as follows: since for every $m$ there is an $n(m)$ with $I^{n(m)} \subset J_mA$ we see that $J_mA$ is open in $A$, hence closed, hence $\Ker(A \to A_m) = J_mA$, in other words, $A_m = A/J_mA$. This holds for every $m$. \medskip\noindent Next, we pick $m$ with $J_mA \subset I$ (\ref{equation-first}). Then choose $n = n(m)$ and finitely generated ideal $J \subset J_m$ with $I^n \subset JA \subset I$ as above. For every $k \geq 1$ we define $\mathfrak b_k = \Ker(B \to A/J^kA)$. Observe that $\mathfrak b_k \supset \mathfrak b_{k + 1}$. For every $k$ there exists an $m'$ with $J_{m'} \subset \mathfrak b_k$ as we have $I^{nk} \subset J^kA$ and we can apply (\ref{equation-first}). On the other hand, for every $m'$ there exists a $k$ such that $I^k \subset J_{m'}A$ because $J_{m'}A$ is open. Then $\mathfrak b_k$ maps to zero in $A/J_{m'}A$ which is faithfully flat over $B/J_{m'}$. Hence $\mathfrak b_k \subset J_{m'}$. In other words, we see that the topology on $B$ is defined by the sequence of ideals $\mathfrak b_k$. Note that $J^k \subset \mathfrak b_k$ which implies that $\mathfrak b_k A = J^kA$. In other words, we have reduced the problem to the situation discussed in the following paragraph. \medskip\noindent We are given a ring map $B \to A$ where \begin{enumerate} \item $B$ is a weakly admissible topological ring with a fundamental system $J_1 \supset J_2 \supset J_3 \supset \ldots$ of ideals of definition, \item $A$ is a ring complete with respect to a finitely generated ideal $I$, \item we have $J_k A = I^k$ for all $k$, and \item $B/J_k \to A/I^k$ is faithfully flat. \end{enumerate} We want to deduce that $B$ is adic*. Pick $g_1, \ldots, g_r \in J_1$ whose images in $A/I^2$ generate $I/I^2$; this is possible because $J_1A/J_2A = I/I^2$. Then for all $k \geq 1$ we see that the elements $g^E = g_1^{e_1} \ldots g_r^{e_r}$ with $|E| = k$ are in $J_k$ because $B/J_k \to A/I^k$ is faithfully flat and $J_1A = I$. Also we have $J_1 J_k \subset J_{k + 1}$ by similar reasoning. The classes of $g^E$ with $|E| = k$ in $J_k/J_{k + 1}$ map to generators of $I^k/I^{k + 1}$ because the images of $g_1, \ldots, g_r$ generate $I/I^2$. Since $B/J_{k + 1} \to A/I^{k + 1}$ is flat we see that $$ J_k/J_{k + 1} \otimes_{B/J_1} A/I = J_k/J_{k + 1} \otimes_{B/J_{k + 1}} A/I^{k + 1} \to I^k/I^{k + 1} $$ is an isomorphism (see More on Morphisms, Lemma \ref{more-morphisms-lemma-deform}). Since $B/J_1 \to A/I$ is faithfully flat, we conclude that the classes of the elements $g^E$, $|E| = k$ generate $J_k/J_{k + 1}$. We claim that $J_k = (g^E, |E| = k)$. Namely, suppose that $x_k \in J_k$. By the above we can write $$ x_k = \sum\nolimits_{|E| = k} b_{E, 0} g^E + x_{k + 1} $$ with $x_{k + 1} \in J_{k + 1}$ and some $b_{E, 0} \in B$. Now we can write $x_{k + 1}$ as follows $$ x_{k + 1} = \sum\nolimits_{|E| = k} \left(\sum\nolimits_{|E'| = 1} b_{E, E'}g^{E'}\right) g^E + x_{k + 2} $$ because every multi-index of degree $k + 1$ is a sum of a multi-index of degree $k$ and a multi-index of degree $1$. Continuing in this manner we can find $b_{E, E'} \in B$ such that for every $l > 1$ we have $$ x_k = \sum\nolimits_{|E| = k} \left(\sum\nolimits_{0 \leq |E'| < l} b_{E, E'} g^{E'}\right) g^E + x_{k + l} $$ with some $x_{k + l} \in J_{k + l}$. Then we can finally define $$ b_E = \sum\nolimits_{E'} b_{E, E'} g^{E'} $$ as an element in $B$ and we see that $x_k = \sum b_E g^E$ as desired. This finishes the proof as now $J_1$ is finitely generated and $J_k = J_1^k$ for all $k \geq 1$.