Let $A\subseteq N\triangleleft G$. Denote $A^c$ the normal closure of $A$ (intersection of all normal subgroups of $G$ containing $A$) and $\langle A\rangle$ the subgroup generated by $A$ (intersection of all subgroups of $G$ containing $A$). We can prove that $\langle A\rangle=N$ implies $A^c=N$. Can we prove that $A^c=N$ implies $\langle A\rangle=N$? What if $G$ is free?
Normal closure of $A$ vs $\langle A\rangle$
group-theory
Related Solutions
It looks fine. But since you point out that $\langle H\rangle$ is the smallest subgroup containing $H$, we can argue more easily (basically the same):
$H$ is the smallest set containing $H$, and it is a subgroup, so it is the smallest subgroup containing $H$.
Hint: To show $\langle S \rangle \subseteq H$, you need to show that $H$ is a subgroup of $G$ containing $S$. To do so, show that the intersection of subgroups is again a subgroup. To see that $H \subseteq \langle S \rangle$, note that $H$ is the intersection of all subgroups of $G$ containing $S$, and that $\langle S \rangle$ is one such subgroup.
Edit: You write in your comment to amrsa that you have shown that the intersection of all subgroups of $G$ containing $S$ contains S. This is true, but not what you need here. Show that for an arbitrary collection $\{K_i\}$ of subgroups of $G$, their intersection $\bigcap_{i \in I} K_i$ is a subgroup of $G$.
For $H \subseteq \langle S \rangle$, if you sit with the definition of $H$ for a bit, you should see that $H \subseteq \langle S \rangle$ is immediate. Think of the case where we have two sets, it's certainly true that $$ A \cap B \subseteq A \text{ and } A \cap B \subseteq B. $$ If $H$ is the intersection of all subgroups containing $S$ and $\langle S \rangle$ is one such subgroup, then it must be that $H \subseteq \langle S \rangle$.
If you want the proof completely in your notation. Let $A = \{K \leq G \mid S \subseteq K\}$, then $H = \cap_{K \in A} K$. As amrsa mentioned, notice that $\langle S \rangle \in A$ and as such $H \subseteq \langle S \rangle$ is immediate because $\langle S \rangle$ is an element of the intersection defining $H$.
Best Answer
General case. No. Let $A$ be a subgroup of $G$ that is not normal. Then $\langle A\rangle=A\neq A^c$.
Free groups. No. Let $G$ be a free group generated by a set $S$ such that $|S|\geq 2$. It is sufficient to show that not all subgroups of $G$ are normal. Indeed, if $x$, $y$ are distinct elements of $S$, then $xy\in x\langle y\rangle$ and $xy\notin \langle y\rangle x$.
Abelian groups. Yes. Obviously, $\langle A\rangle\subseteq A^c$. But we also have $A^c\subseteq\langle A\rangle$ since $\langle A\rangle$ is normal and $A\subseteq\langle A\rangle$.