Proof.
Part (1) is Lemma 87.37.1. If X is McQuillan, then X = \text{Spf}(A) for some weakly admissible topological ring A. Then X_{/T} \to X \to \mathop{\mathrm{Spec}}(A) satisfies property (2) of Lemma 87.9.6 and hence X_{/T} is McQuillan, see Definition 87.9.7.
Assume X and T are as in (3). Then X = \text{Spf}(A) where A has a fundamental system A \supset I_1 \supset I_2 \supset I_3 \supset \ldots of weak ideals of definition, see Lemma 87.10.4. By Algebra, Lemma 10.29.1 we can find a finitely generated ideal \overline{J} = (\overline{f}_1, \ldots , \overline{f}_ r) \subset A/I_1 such that T is cut out by \overline{J} inside \mathop{\mathrm{Spec}}(A/I_1) = |X_{red}|. Choose f_ i \in A lifting \overline{f}_ i. If Z = \mathop{\mathrm{Spec}}(B) is an affine scheme and g : Z \to X is a morphism with g(Z) \subset T (set theoretically), then g^\sharp : A \to B factors through A/I_ n for some n and g^\sharp (f_ i) is nilpotent in B for each i. Thus J_{m, n} = (f_1, \ldots , f_ r)^ m + I_ n maps to zero in B for some n, m \geq 1. It follows that X_{/T} is the formal spectrum of \mathop{\mathrm{lim}}\nolimits _{n, m} A/J_{m, n} and hence countably indexed. This proves (3).
Proof of (4). Here the argument is the same as in (3). However, here we may choose I_ n = I^ n for some finitely generated ideal I \subset A. Then it is clear that X_{/T} is the formal spectrum of \mathop{\mathrm{lim}}\nolimits A/J^ n where J = (f_1, \ldots , f_ r) + I. Some details omitted.
Proof of (5). In this case X_{red} is the spectrum of a Noetherian ring and hence the assumption that |X_{red}| \setminus T is quasi-compact is satisfied. Thus as in the proof of (4) we see that X_{/T} is the spectrum of \mathop{\mathrm{lim}}\nolimits A/J^ n which is a Noetherian adic topological ring, see Algebra, Lemma 10.97.6.
\square
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