In this question Prove that the product $NK$ of two normal subgroups $N$ and $K$ of a group $G$ is a normal subgroup of $G$, and $NK=KN$., it is proved that $NK=KN$ if both $N$ and $K$ are normal subgroups of a group $G$. I think in Artin's Algebra it is proved with only one of them assumed to be normal but both are still subgroups.
Because $N$ is normal: $\forall g \in G, \forall n \in N, gng^{-1} \in N$
$$\implies \forall k \in K, \forall n \in N, knk^{-1} \in N$$
$$\implies \forall k \in K, \forall n \in N, kn \in Nk$$
$$\implies \forall k \in K, kN \subseteq Nk$$
$$KN \subseteq NK$$
Also: $\forall g \in G, \forall n \in N, g^{-1}ng \in N$
$$\implies \forall k \in K, \forall n \in N, k^{-1}nk \in N$$
$$\implies \forall k \in K, \forall n \in N, nk \in kN$$
$$\implies \forall k \in K, Nk \subseteq kN$$
$$NK \subseteq KN$$
Do I correctly prove that $NK=KN$ above and also understand the proof in Artin's Algebra? The following is a screenshot of a proof: (Kiefer Sutherland's voice)
Your proof is fine and you don't even need to assume that $K$ is a subgroup (you haven't used it anywhere).