Let $T \subset U$ be the set of points $x$ with $\overline{\{x\}} \cap V(I) \subset V(\mathfrak a)$. Let $\mathcal{F} \to \mathcal{F}'$ be the surjection of coherent modules on $U$ constructed in Local Cohomology, Lemma \ref{local-cohomology-lemma-get-depth-1-along-Z}. Since $\mathcal{F} \to \mathcal{F}'$ is an isomorphism over an open $V \subset U$ containing $U \cap V(I)$ it suffices to prove the lemma with $\mathcal{F}$ replaced by $\mathcal{F}'$. Hence we may and do assume for $x \in U$ with $\overline{\{x\}} \cap V(I) \subset V(\mathfrak a)$ we have $\text{depth}(\mathcal{F}_x) \geq 1$. \medskip\noindent Let $\mathcal{V}$ be the set of open subschemes $V \subset U$ containing $U \cap V(I)$ ordered by reverse inclusion. This is a directed set. We first claim that $$ \mathcal{F}(V) \longrightarrow \lim H^0(U, \mathcal{F}/I^n\mathcal{F}) $$ is injective for any $V \in \mathcal{F}$ (and in particular the map of the lemma is injective). Namely, an associated point $x$ of $\mathcal{F}$ must have $\overline{\{x\}} \cap U \cap Y \not = \emptyset$ by the previous paragraph. If $y \in \overline{\{x\}} \cap U \cap Y$ then $\mathcal{F}_x$ is a localization of $\mathcal{F}_y$ and $\mathcal{F}_y \subset \lim \mathcal{F}_y/I^n \mathcal{F}_y$ by Krull's intersection theorem (Algebra, Lemma \ref{algebra-lemma-intersect-powers-ideal-module-zero}). This proves the claim as a section $s \in \mathcal{F}(V)$ in the kernel would have to have empty support, hence would have to be zero. \medskip\noindent Choose a finite $A$-module $M$ such that $\mathcal{F}$ is the restriction of $\widetilde{M}$ to $U$, see Local Cohomology, Lemma \ref{local-cohomology-lemma-finiteness-pushforwards-and-H1-local}. We may and do assume that $H^0_\mathfrak a(M) = 0$. Let $\text{Ass}(M) \setminus V(I) = \{\mathfrak p_1, \ldots, \mathfrak p_n\}$. We will prove the lemma by induction on $n$. After reordering we may assume that $\mathfrak p_n$ is a minimal element of the set $\{\mathfrak p_1, \ldots, \mathfrak p_n\}$ with respect to inclusion, i.e, $\mathfrak p_n$ is a generic point of the support of $M$. Set $$ M' = H^0_{\mathfrak p_1 \ldots \mathfrak p_{n - 1} I}(M) $$ and $M'' = M/M'$. Let $\mathcal{F}'$ and $\mathcal{F}''$ be the coherent $\mathcal{O}_U$-modules corresponding to $M'$ and $M''$. Dualizing Complexes, Lemma \ref{dualizing-lemma-divide-by-torsion} implies that $M''$ has only one associated prime, namely $\mathfrak p_n$. On the other hand, since $\mathfrak p_n \not \in V(\mathfrak p_1 \ldots \mathfrak p_{n - 1} I)$ we see that $\mathfrak p_n$ is not an associated prime of $M'$. Hence the induction hypothesis applies to $M'$; note that since $\mathcal{F}' \subset \mathcal{F}$ the condition $\text{depth}(\mathcal{F}'_x) \geq 1$ at points $x$ with $\overline{\{x\}} \cap V(I) \subset V(\mathfrak a)$ holds, see Algebra, Lemma \ref{algebra-lemma-depth-in-ses}. \medskip\noindent Let $\hat s$ be an element of $\lim H^0(U, \mathcal{F}/I^n\mathcal{F})$. Let $\hat s''$ be the image in $\lim H^0(U, \mathcal{F}''/I^n\mathcal{F}'')$. Since $\mathcal{F}''$ has only one associated point, namely the point corresponding to $\mathfrak p_n$, we see that Lemma \ref{lemma-application-H0-pre} applies and we find an open $U \cap V(I) \subset V \subset U$ and a section $s'' \in \mathcal{F}''(V)$ mapping to $\hat s''$. Let $J \subset A$ be an ideal such that $V(J) = \Spec(A) \setminus V$. By Cohomology of Schemes, Lemma \ref{coherent-lemma-homs-over-open} after replacing $J$ by a power, we may assume there is an $A$-linear map $\varphi : J \to M''$ corresponding to $s''$. Since $M \to M''$ is surjective, for each $g \in J$ we can choose $m_g \in M$ mapping to $\varphi(g) \in M''$. Then $\hat s'_g = g \hat s - m_g$ is in $\lim H^0(U, \mathcal{F}'/I^n\mathcal{F}')$. By induction hypothesis there is a $V' \geq V$ section $s'_g \in \mathcal{F}'(V')$ mapping to $\hat s'_g$. All in all we conclude that $g \hat s$ is in the image of $\mathcal{F}(V') \to \lim H^0(U, \mathcal{F}/I^n\mathcal{F})$ for some $V' \subset V$ possibly depending on $g$. However, since $J$ is finitely generated we can find a single $V' \in \mathcal{V}$ which works for each of the generators and it follows that $V'$ works for all $g$. \medskip\noindent Combining the previous paragraph with the injectivity shown in the second paragraph we find there exists a $V' \geq V$ and an $A$-module map $\psi : J \to \mathcal{F}(V')$ such that $\psi(g)$ maps to $g\hat s$. This determines a map $\widetilde{J} \to (V' \to \Spec(A))_*\mathcal{F}|_{V'}$ whose restriction to $V'$ provides an element $s \in \mathcal{F}(V')$ mapping to $\hat s$. This finishes the proof.