I think I have somewhat of a roundabout way to show that the commutator subgroup $[G,G]$ is normal in $G$. Please check my proof for errors/improvements.
Let $G$ be a group and define $S := \{xyx^{-1}y^{-1}: x,y \in G\}$, then $$[G,G] = \langle S\rangle = \bigcap_{S \subseteq H < G} H, $$ so it is clear that $[G,G]$ is at least a subgroup. For normality consider the group action homomorphism $$ \psi \colon G \to \text{Perm}(\langle S \rangle), \quad g \mapsto \phi_g $$ where $\phi_g \colon \langle S \rangle \to \langle S \rangle $ is defined by $ s \mapsto g^{-1} s g$. The conjugation map is a homomorphism (i.e. $\phi_g$ is a homomorphism for every $g \in G$), and in this case is an automorphism so $\langle S \rangle$ is normal in $G$.
This is a bit different than the "standard" way of showing normality (i.e. via the definition), so I just wanted to make sure my reasoning is correct.
This is a good effort, but there's an issue to address. By considering the homomorphism
$$\psi\colon G \to \operatorname{Perm}(\langle S \rangle),$$ you are tacitly assuming that $\langle S \rangle = [G,G]$ is normal. More specifically: the statement that for all $g$ in $G$, $\phi_g$ takes elements of $\langle S \rangle$ to elements of $\langle S \rangle$ is equivalent to the statement that $\langle S \rangle$ is normal.
However, you've come close to one nice way of showing that $[G,G]$ is normal. Some general set-up: suppose $H = \langle a_i \rangle_{i\in I} \subset G$ is a subgroup. To show that $H$ is normal, it's sufficient to show that $g a_i g^{-1} \in H$ for all $i \in I$ and $g \in G$.
We can apply this to the generating set $S$ that you define: given $xyx^{-1}y^{-1} \in S$,
$$g(xyx^{-1}y^{-1})g^{-1}=(gxg^{-1})(gyg^{-1})(gx^{-1}g^{-1})(gy^{-1}g^{-1})$$ $$= (gxg^{-1})(gyg^{-1})(gxg^{-1})^{-1}(gyg^{-1})^{-1} \in \langle S \rangle.$$
This is essentially the same argument that you need to give to show that map $\psi \colon G\to\operatorname{Perm}([G,G])$ is well-defined.