The Problem:
Let $G$ be a finite group and $p$ a prime divisor of $|G|$.
Prove that a normal $p$-subgroup of $G$ is contained in every $p$-Sylow subgroup of $G$.
My Attempt(s):
Here's a slight modification of an argument I found elsewhere (which I have an issue with): Let $Q \in \mathscr{S}$, where $\mathscr{S}$ is the set of $p$-Sylow subgroups of $G$. Consider the set $A = \bigcap_{g \in G} gQg^{-1}$. Clearly $A \subset Q$ (since $A \subset eQe^{-1} = Q$); so we need just need to show that $A \triangleleft G$. However, it's not entirely clear to me that $A$ is necessarily a $p$-subgroup of $G$, since (it seems to me) it's possible that $A$ might be trivial (where I assume that groups of order 1 are not considered $p$-groups).
If my suspicion is wrong, then the remainder of the proof is fine
I don't see the point of considering $A$. You need to prove this for any normal $p-$subgroup of $G$, not just for $A$. Indeed, you can show that $A$ contains every normal $p-$subgroup of $G$, however this doesn't seem easier than the original problem.
What I would do is the following. Let $N$ be a normal $p-$subgroup of $G$. Then by the Sylow's Theorems we have that it is contained in a Sylow $p-$subgroup $P$ of $G$, i.e. $N \subseteq P$. Now by the Sylow's Theorem if $Q$ is any Sylow $p-$subgroup we have that $Q = g^{-1}Pg$ for some $g \in G$. So finally:
$$N \subseteq P \implies N = g^{-1}Ng \subseteq g^{-1}Pg = Q$$
Thus $N$ is contained in every Sylow $p-$subgroup of $G$.