I'm having trouble proving that if $H$ is a non-normal subgroup of a group $G$, then there exists $g \in G$ such that $HgH$ is not a left coset. Here is my attempt, some steps of which I'm not totally sure of (and which I've bolded).
We proceed by contraposition. Suppose that for every $g \in G$, the double coset $HgH$ is a left coset; we will prove that $H$ is normal. Fix $g \in G$. I claim that $HgH = gH$. Indeed, if $x \in gH$, then $x = gh$ for some $h \in H$, hence $x = ghe \in HgH$ since $H$ is a subgroup and therefore contains the identity. Therefore, we have $HgH \supset gH$. As $HgH$ must equal some left coset, and the left cosets partition $G$ into equivalence classes, which are either equal or disjoint, we must have $HgH = gH$.
We will show that $H$ is normal, i.e., that $Hg = gH$. If $x \in Hg$, then $x = hge$ for some $h \in H$, hence $x \in HgH$, hence $x \in gH$, so $Hg \subset gH$. Multiplying by $g^{-1}$ on the left, we find that $g^{-1} Hg \subset H$. Conversely, multiplying by $g$ on the left and $H$ on the right gives $H \subset gHg^{-1}$. As this holds for all $g \in G$, it holds upon replacing $g$ with $g^{-1}$, hence $H \subset g^{-1} H g$, so $H = g^{-1} H g$, hence $gH = Hg$, so $H$ is normal.
I'd appreciate some help with the bolded parts of my attempted proof above. I'm especially not certain on the argument for why $HgH = gH$, but I couldn't figure out how to prove directly that $H g H \subset gH$.
For your bolded part:
Since $HgH$ is some left coset of $H$, let $HgH=aH$, you have shown $gH\subseteq HgH=aH$, hence $$\exists x, ~ x\in gH\land x\in aH\Longrightarrow \exists h_1,h_2\in H,~~ gh_1=x=ah_2$$
then we have
$$a=gh_1h_2^{-1}=gh_3,~~~h_3=h_1h_2^{-1}\in H$$
hence
$$HgH=aH=(gh_3)H=g(h_3H)=gH$$
Next, a quicker way to show $H$ is normal,
$$\forall g\in G, \forall h\in H,~~ \color{red}{g^{-1}hg}=g^{-1}hge \color{red}\in g^{-1}HgH=\color{red}H$$