It turns out to be false. Consider, e.g., the dihedral group $D_8$ of order $8$, which can be viewed as consisting of reflections and rotations. The subgroup $H$ generated by any reflection $h$ of order $2$ is abelian but not a normal subgroup of $D_8$. Indeed, if $g\in D_8$ has order $4$, then $g^{-1}Hg$ is a different subgroup of order $2$.
More generally, here is a necessary and sufficient condition for an order $2$ subgroup to be a normal subgroup.
Proposition. Let $G$ be any group. If $N$ is a subgroup of $G$ such that $|N|=2$, then $N\lhd G$ if and only if $N\subseteq Z(G)$, where $Z(G)$ is the center of $G$.
Proof. The backward implication follows from the fact that conjugation by any element of $G$ fixes any element of $Z(G)$, including every element of $N$.
We turn, now, to the forward implication. Let $N\lhd G$, and suppose $N=\{e,x\}$, where $e$ is the identity element of $G$, and $x$ is any non-identity element of $G$. We note that $e\in Z(G)$, because $Z(G)$ is a subgroup of $G$. Next, let $g\in G$. We have,
$$g^{-1}xg\in g^{-1}Ng=N$$
because $N$ is a normal subgroup of $G$. Thus, $g^{-1}xg=x$ or $g^{-1}xg=e$. If $g^{-1}xg=e$, then $x=gg^{-1}=e$, a contradiction. Hence, $g^{-1}xg = x$. It follows that $x$ commutes with every element of $G$, so $x\in Z(G)$. Since $e$ and $x$ both belong to $Z(G)$, we have $N\subseteq Z(G)$. q.e.d.
