Let us consider the set $$ B = \{\mathfrak p \not \in V(I),\ \mathfrak p \in T,\text{ and } \text{depth}(M_\mathfrak p) \leq s\} $$ Choose $J_0 \subset J$ such that $V(J_0)$ is the closure of $B \cup V(J)$. \medskip\noindent Claim I: $V(J_0) \cap V(I) = V(J) \cap V(I)$. \medskip\noindent Proof of Claim I. The inclusion $\supset$ holds by construction. Let $\mathfrak p$ be a minimal prime of $V(J_0)$. If $\mathfrak p \in B \cup V(J)$, then either $\mathfrak p \in T$ or $\mathfrak p \in V(J)$ and in both cases $V(\mathfrak p) \cap V(I) \subset V(J) \cap V(I)$ as desired. If $\mathfrak p \not \in B \cup V(J)$, then $V(\mathfrak p) \cap B$ is dense, hence infinite, and we conclude that $\text{depth}(M_\mathfrak p) < s$ by Local Cohomology, Lemma \ref{local-cohomology-lemma-depth-function}. In fact, let $V(\mathfrak p) \cap B = \{\mathfrak p_\lambda\}_{\lambda \in \Lambda}$. Pick $\mathfrak q_\lambda \in V(\mathfrak p_\lambda) \cap V(J) \cap V(I)$ as in (3). Let $\delta : \Spec(A) \to \mathbf{Z}$ be the dimension function associated to a dualizing complex $\omega_A^\bullet$ for $A$. Since $\Lambda$ is infinite and $\delta$ is bounded, there exists an infinite subset $\Lambda' \subset \Lambda$ on which $\delta(\mathfrak q_\lambda)$ is constant. For $\lambda \in \Lambda'$ we have $$ \text{depth}(M_{\mathfrak p_\lambda}) + \delta(\mathfrak p_\lambda) - \delta(\mathfrak q_\lambda) = \text{depth}(M_{\mathfrak p_\lambda}) + \dim((A/\mathfrak p_\lambda)_{\mathfrak q_\lambda}) \leq d + s $$ by (3) and the definition of $B$. By the semi-continuity of the function $\text{depth} + \delta$ proved in Duality for Schemes, Lemma \ref{duality-lemma-sitting-in-degrees} we conclude that $$ \text{depth}(M_\mathfrak p) + \dim((A/\mathfrak p)_{\mathfrak q_\lambda}) = \text{depth}(M_\mathfrak p) + \delta(\mathfrak p) - \delta(\mathfrak q_\lambda) \leq d + s $$ Since also $\mathfrak p \not \in V(I)$ we read off from (4) that $\mathfrak p \in T$, i.e., $V(\mathfrak p) \cap V(I) \subset V(J) \cap V(I)$. This finishes the proof of Claim I. \medskip\noindent Claim II: $H^i_{J_0}(M) \to H^i_J(M)$ is an isomorphism for $i \leq s$ and $J' \subset J_0$ with $V(J') \cap V(I) = V(J) \cap V(I)$. \medskip\noindent Proof of claim II. Choose $\mathfrak p \in V(J')$ not in $V(J_0)$. It suffices to show that $H^i_{\mathfrak pA_\mathfrak p}(M_\mathfrak p) = 0$ for $i \leq s$, see Local Cohomology, Lemma \ref{local-cohomology-lemma-isomorphism}. Observe that $\mathfrak p \in T$. Hence since $\mathfrak p$ is not in $B$ we see that $\text{depth}(M_\mathfrak p) > s$ and the groups vanish by Dualizing Complexes, Lemma \ref{dualizing-lemma-depth}. \medskip\noindent Claim III. The final statement of the lemma is true. \medskip\noindent By Claim II for $i \leq s$ we have $$ H^i_T(M) = H^i_{J_0}(M) = H^i_{J'}(M) $$ for all ideals $J' \subset J_0$ with $V(J')  \cap V(I) = V(J) \cap V(I)$. See Local Cohomology, Lemma \ref{local-cohomology-lemma-adjoint-ext}. Let us check the hypotheses of Local Cohomology, Proposition \ref{local-cohomology-proposition-annihilator} for the subsets $T \subset T \cup V(I)$, the module $M$, and the integer $s$. We have to show that given $\mathfrak p \subset \mathfrak q$ with $\mathfrak p \not \in T \cup V(I)$ and $\mathfrak q \in T$ we have $$ \text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim((A/\mathfrak p)_\mathfrak q) > s $$ If $\text{depth}(M_\mathfrak p) \geq s$, then this is true because the dimension of $(A/\mathfrak p)_\mathfrak q$ is at least $1$. Thus we may assume $\text{depth}(M_\mathfrak p) < s$. If $\mathfrak q \in V(I)$, then $\mathfrak q \in V(J) \cap V(I)$ and the inequality holds by (4). If $\mathfrak q \not \in V(I)$, then we can use (3) to pick $\mathfrak q' \in V(\mathfrak q) \cap V(J) \cap V(I)$ with $\dim((A/\mathfrak q)_{\mathfrak q'}) \leq d$. Then assumption (4) gives $$ \text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim((A/\mathfrak p)_{\mathfrak q'}) > s + d $$ Since $A$ is catenary this implies the inequality we want. Applying Local Cohomology, Proposition \ref{local-cohomology-proposition-annihilator} we find $J'' \subset A$ with $V(J'') \subset T \cup V(I)$ such that $J''$ annihilates $H^i_T(M)$ for $i \leq s$. Then we can write $V(J'') \cup V(J_0) \cup V(I) = V(J'I)$ for some $J' \subset J_0$ with $V(J') \cap V(I) = V(J) \cap V(I)$. Replacing $J_0$ by $J'$ the proof is complete.