Let $H$ be a $p$-subgroup of a finite group $G$, then $H$ is contained in a Sylow $p$-subgroup of $G$. So this is a more general case of a well-known problem where $H$ is a normal $p$-subgroup of $G$, then $H$ is contained in every Sylow $p$-subgroup.
So, now, we do not have the option that it is normal.
I tried the same approach as when it is normal where I would the Second Sylow Theorem and show the containment within another group. But this doesn't work anymore because it is not normal anymore.
I also tried a different approach where $H$ is a fixed subgroup of $G$ and $A, B$ are any subgroups of $G$, a H-conjugacy class where $x\in H$, $B = x^{-1}Ax=\{x^{-1}ax:a \in A\}$ 
I thought I was going somewhere with the $H$-conjugacy class..
Would you provide a proof please?
 A: Let $P \in Syl_p(G)$ and $X=\{Px : x \in G \}$ the set of right cosets of $P$ in $G$ and observe that $|X|=|G:P|$ is not divisible by $p$ since $P$ is a Sylow $p$-subgroup. Now let act $H$ on $X$ by right multiplication. Since the lengths of the orbits divide the $p$-power $|H|$, and $|X|$ is not divisible by $p$, there must be at least one orbit of length $1$, say the orbit of $Pg$. Hence $Pgh=Pg$ for all $h \in H$, which is equivalent to $H \subseteq P^g$. And of course, $P^g$ is a Sylow $p$-subgroup when $P$ is one.
A: Could I suggest a less technical proof? For that let's define a Sylow $p$-subgroup as a maximal $p$-subgroup of $G$, i.e. $P$ is a Sylow $p$-subgroup of $G$ if for a $p$-subgroup $K$ of $G$ with $P \subset K$, it follows $P = K$.
Now if $H$ is $p$-subgroup of $G$ it is either properly contained in a $p$-subgroup or not. If it is not properly contained in a $p$-subgroup it is a Sylow $p$-subgroup by definition. Otherwise let $H^{\prime}$ be a $p$-subgroup which properly containes $H$, i.e. $H\subsetneq H^{\prime}$. Continuing in this way we get a sequence of $p$-subgroups:
$$H\subsetneq H^{\prime} \subsetneq H^{\prime \prime} \cdots,$$ which has to terminate since $G$ is finite.
Remark: Using Zorn's Lemma the same proof applies to infinite $G$.
