The (left) cosets of $H$ partition the group $G$; since there are two cosets and one of them is $H$, that means that, whatever the other one is (call it $\mathcal{C}$), as a set, you have that $H\cap\mathcal{C}=\emptyset$ and $H\cup
\mathcal{C}=G$. These two together tell you that the other one must be equal, as a set, to the complement of $H$ inside of $G$; that is, $\mathcal{C}=G-H$. So the two cosets are $H$ and $G-H$. Of course, $G-H$ can be written as $xH$ for some $x$; in fact, for any $x\notin H$ you have $xH = G-H$.
But the exact same argument holds for right cosets: one of them is $H$, and the other one must be, as a set, the complement of $H$. So the two right cosets are $H$ and $G-H$, and $G-H$ can be written as $Hy$ for some $y$; in fact, for any $y\notin H$ you have $Hy=G-H$.
Now note that in both cases, the cosets are:
1. $H$, the coset of the elements of $H$; and
2. $G-H$, the coset of the elements not in $H$.
So each of the two left cosets is also a right coset and vice-versa.
Choose $h\in H$ and $g\in G$. Then $gH = gh^{-1}H$ trivially, so $Hg = Hgh^{-1}$ by the hypothesis. So by the definition of these sets being equal there is some $h'\in H$ so that $g = h'gh^{-1}$, which is equivalent to $ghg^{-1} = h' \in H$.
And this exactly means that $gHg^{-1} = H$ for every $g\in G$.
Best Answer
Consider $aga^{-1}$ for $g\in H$ arbitrary. Then $ag\in aH=Ha$, so $ag=g^{\prime}a$ for some $g^{\prime}\in H$. Hence, $aga^{-1}=g^{\prime}$, and we get the result you need, that $aHa^{-1}\subseteq H$.
You should realise that this all implies that the given condition implies that the subgroup $H$ is a normal subgroup of $G$.
A different idea to the proof you are proposing would be to simply note the following. $$aHa^{-1}=H\Leftrightarrow aHa^{-1}a=Ha\Leftrightarrow aH=Ha$$ This is slicker, but I understand why you would not necessarily be comfortable with it.
(Small point: You write $\subseteq$, as if the things we are dealing with are just subsets. However, they are subgroups, so $\leq$ would be more appropriate. Challenge: Prove what I just said, that $aHa^{-1}\leq G$ for all $H\leq G$ and all $a\in G$.)