Minimal normal subgroups of normal closure

This question arose from my attempt at understanding the answer in this post - the comments below it, to be precise. Everything revolves around the following problem:

Let $$S$$ be a simple, nonabelian and subnormal subgroup of a group $$G$$. Show that $$S^G$$, the normal closure of $$S$$ in $$G$$ is a minimal normal subgroup of $$G$$ .

The question (from Isaacs' book "Finite Group Theory") comes with a hint. It says:

HINT: Work by induction on $$|G|$$ to conclude $$S \subset \operatorname{Soc}(H)$$ whenever $$S \subset H$$. Deduce that each conjugate of $$S$$ in $$G$$ is a minimal normal subgroup of $$S^G$$. Then, apply the previous problem to $$S^G$$.

I managed to do everything, except for the part in italics. What I did was (briefly) as follows:

1. $$S \lhd \lhd G \implies S \lhd \lhd S^G$$. If $$S^G = G$$, then $$S = G$$, which means the result is trivial. Therefore, we can assume $$S^G \neq G$$
2. Using induction, it was simple to prove $$S \subset \operatorname{Soc}(H)$$ if $$H \neq G$$. In particular, $$S \subset \operatorname{Soc}(S^G) \implies S^G = \operatorname{Soc}(S^G)$$
3. Using a previous problem, $$S^G$$ is a direct product of minimal normal subgroups (and of simple groups)

In the aforementioned post, user @Stefan4024 stated:

Since $$S \lhd \lhd S^G$$, and $$S^G$$ is a product of minimal normal subgroups, then $$S$$ is a minimal normal subgroup of $$S^G$$ (I took $$K = S^G$$)

I just can’t see why this follows so immediately.

Continuing my previous reasoning, I managed to find, by subnormality, a normal subgroup $$S \lhd \lhd K \lhd S^G$$ (which is non-abelian) and this yielded, by Exercise $$2A.6$$, a minimal normal subgroup $$X$$ of $$S^G$$ such that $$X \subset K$$. It follows, both by simplicity and from the fact that minimal normal subgroups normalize all subnormal subgroups, that either $$S = X$$ or $$S \cap X = 1$$. And I couldn't rule out this last case...

Is there a simpler way to prove the statement in the last block quote? If not, how do I continue from my arguments?

Since $$S$$ is simple and subnormal in $$S^G$$, $$S$$ must be one of these direct factors, so it is indeed a minimal normal subgroup of $$S^G$$.