The assumption that $X_{/T}$ is countably indexed means that there exists a sequence of ideals $$ A \supset J_1 \supset J_2 \supset J_3 \supset \ldots $$ with $V(J_n) = T$ such that every ideal $J \subset A$ with $V(J) = T$ there exists an $n$ such that $J \supset J_n$. \medskip\noindent To construct an example for (2) let $\omega_1$ be the first uncountable ordinal. Let $k$ be a field and let $A$ be the $k$-algebra generated by $x_\alpha$, $\alpha \in \omega_1$ and $y_{\alpha \beta}$ with $\alpha \in \beta \in \omega_1$ subject to the relations $x_\alpha = y_{\alpha \beta} x_\beta$. Let $T = V(x_\alpha)$. Let $J_n = (x_\alpha^n)$. If $J \subset A$ is an ideal such that $V(J) = T$, then $x_\alpha^{n_\alpha} \in J$ for some $n_\alpha \geq 1$. One of the sets $\{\alpha \mid n_\alpha = n\}$ must be unbounded in $\omega_1$. Then the relations imply that $J_n \subset J$. \medskip\noindent To see that (2) holds it now suffices to show that $A^\wedge = \lim A/J_n$ is not a ring complete with respect to a finitely generated ideal. For $\gamma \in \omega_1$ let $A_\gamma$ be the quotient of $A$ by the ideal generated by $x_\alpha$, $\alpha \in \gamma$ and $y_{\alpha \beta}$, $\alpha \in \gamma$. As $A/J_1$ is reduced, every topologically nilpotent element $f$ of $\lim A/J_n$ is in $J_1^\wedge = \lim J_1/J_n$. This means $f$ is an infinite series involving only a countable number of generators. Hence $f$ dies in $A_\gamma^\wedge = \lim A_\gamma/J_nA_\gamma$ for some $\gamma$. Note that $A^\wedge \to A_\gamma^\wedge$ is continuous and open by Lemma \ref{lemma-ses}. If the topology on $A^\wedge$ was $I$-adic for some finitely generated ideal $I \subset A^\wedge$, then $I$ would go to zero in some $A_\gamma^\wedge$. This would mean that $A_\gamma^\wedge$ is discrete, which is not the case as there is a surjective continuous and open (by Lemma \ref{lemma-ses}) map $A_\gamma^\wedge \to k[[t]]$ given by $x_\alpha \mapsto t$, $y_{\alpha \beta} \mapsto 1$ for $\gamma = \alpha$ or $\gamma \in \alpha$. \medskip\noindent Before we prove (1) we first prove the following: If $I \subset A^\wedge$ is a finitely generated ideal whose closure $\bar I$ is open, then $I = \bar I$. Since $V(J_n^2) = T$ there exists an $m$ such that $J_n^2 \supset J_m$. Thus, we may assume that $J_n^2 \supset J_{n + 1}$ for all $n$ by passing to a subsequence. Set $J_n^\wedge = \lim_{k \geq n} J_n/J_k \subset A^\wedge$. Since the closure $\bar I = \bigcap (I + J_n^\wedge)$ (Lemma \ref{lemma-closed}) is open we see that there exists an $m$ such that $I + J_n^\wedge \supset J_m^\wedge$ for all $n \geq m$. Fix such an $m$. We have $$ J_{n - 1}^\wedge I + J_{n + 1}^\wedge \supset J_{n - 1}^\wedge (I + J_{n + 1}^\wedge) \supset J_{n - 1}^\wedge J_m^\wedge $$ for all $n \geq m + 1$. Namely, the first inclusion is trivial and the second was shown above. Because $J_{n - 1}J_m \supset J_{n - 1}^2 \supset J_n$ these inclusions show that the image of $J_n$ in $A^\wedge$ is contained in the ideal $J_{n - 1}^\wedge I + J_{n + 1}^\wedge$. Because this ideal is open we conclude that $$ J_{n - 1}^\wedge I + J_{n + 1}^\wedge \supset J_n^\wedge. $$ Say $I = (g_1, \ldots, g_t)$. Pick $f \in J_{m + 1}^\wedge$. Using the last displayed inclusion, valid for all $n \geq m + 1$, we can write by induction on $c \geq 0$ $$ f = \sum f_{i, c} g_i \mod J_{m + 1+ c}^\wedge $$ with $f_{i, c} \in J_m^\wedge$ and $f_{i, c} \equiv f_{i, c - 1} \bmod J_{m + c}^\wedge$. It follows that $IJ_m^\wedge \supset J_{m + 1}^\wedge$. Combined with $I + J_{m + 1}^\wedge \supset J_m^\wedge$ we conclude that $I$ is open. \medskip\noindent Proof of (1). Assume $T = V(f_1, f_2, f_3, \ldots)$. Let $I_m \subset A^\wedge$ be the ideal generated by $f_1, \ldots, f_m$. We distinguish two cases. \medskip\noindent Case I: For some $m$ the closure of $I_m$ is open. Then $I_m$ is open by the result of the previous paragraph. For any $n$ we have $(J_n)^2 \supset J_{n+1}$ by design, so the closure of $(J_n^\wedge)^2$ contains $J_{n+1}^\wedge$ and thus is open.  Taking $n$ large, it follows that the closure of the product of any two open ideals in $A^\wedge$ is open. Let us prove $I_m^k$ is open for $k \ge 1$ by induction on $k$. The case $k = 1$ is our hypothesis on $m$ in Case I. For $k > 1$, suppose $I_m^{k - 1}$ is open. Then $I_m^k = I_m^{k - 1} \cdot I_m$ is the product of two open ideals and hence has open closure. But then since $I_m^k$ is finitely generated it follows that $I_m^k$ is open by the previous paragraph (applied to $I = I_m^k$), so we can continue the induction on $k$. As each element of $I_m$ is topologically nilpotent, we conclude that $I_m$ is an ideal of definition which proves that $A^\wedge$ is adic with a finitely generated ideal of definition, i.e., $X_{/T}$ is adic*. \medskip\noindent Case II. For all $m$ the closure $\bar I_m$ of $I_m$ is not open. Then the topology on $A^\wedge/\bar I_m$ is not discrete. This means we can pick $\phi(m) \geq m$ such that $$ \Im(J_{\phi(m)} \to A/(f_1, \ldots, f_m)) \not = \Im(J_{\phi(m) + 1} \to A/(f_1, \ldots, f_m)) $$ To see this we have used that $A^\wedge/(\bar I_m + J_n^\wedge) = A/((f_1, \ldots, f_m) + J_n)$. Choose exponents $e_i > 0$ such that $f_i^{e_i} \in J_{\phi(m) + 1}$ for $0 < m < i$. Let $J = (f_1^{e_1}, f_2^{e_2}, f_3^{e_3}, \ldots)$. Then $V(J) = T$. We claim that $J \not \supset J_n$ for all $n$ which is a contradiction proving Case II does not occur. Namely, the image of $J$ in $A/(f_1, \ldots, f_m)$ is contained in the image of $J_{\phi(m) + 1}$ which is properly contained in the image of $J_m$.