Let $G=H \times K$ where $H$ and $K$ are groups, $H_1 \le H, K_1 \le K$. Using the fact that $[(h_1,k_1),(h,k)] = ([h_1,h],[k_1,k])$ I could prove that $$[H_1 \times K_1, H \times K] \subseteq [H_1,H] \times [K_1,K]. \tag{1}$$ But I haven't been able to prove the reverse inclusion.
The difficulty is that the right side in $(1)$ is $\langle [h_1,h]: h_1 \in H_1, h \in H\rangle \times \langle [k_1,k]: k_1 \in K_1, k \in K\rangle$, that is it is of the form $\langle A\rangle \times \langle B\rangle,$ where $A$ is a subset of $H$ and $B$ is a subset of $K$.
How could I prove the reverse inclusion?
Thanks. I have not enough space for my answer in the comments window. Notation: $\langle [h_1,h] \rangle$ will stand for $\langle \{[h_1,h]: h_1 \in H_1, h \in H\} \rangle$. Let $L = [H_1 \times K_1, H \times K]$. You say it is enough to show that $([h_1,h],[k_1,k])$ lies in $L$. But that proves that $\langle ([h_1,h],[k_1,k]) \rangle$ lies in $L$. Instead I need to prove that $\langle [h_1,h] \rangle \times \langle [k_1,k] \rangle$ lies in $L$.
I found two ways to prove the reverse inclusion. (a) To work with the internal direct product. So $G$ is now the internal direct product of its subgroups $H$ and $K$, $G = HK$. And I have to prove that, if $H_1 \le H$ and $K_1 \le K$ (I don't know if this condition is necessary) then $[H_1,H][K_1,K] \subseteq [H_1 K_1,HK] =: L$. But $h_1 \in H_1K_1$ and $h \in HK$. So $[h_1,h] \in L$ for all $h_1$ and $h$ and hence $\langle [h_1,h]\rangle \le L$. Similarly $\langle [k_1,k]\rangle \le L$ and the inclusion is proved.
(b) $\langle [h_1,h]\rangle \times 1 = \langle ([h_1,h],1)\rangle = \langle [(h_1,1),(h,1)]\rangle \le L$. Similarly $1 \times \langle [k_1,k]\rangle \le L$. From which I infer $\langle [h_1,h]\rangle \times \langle [k_1,k]\rangle \le L$. That is $[H_1,H] \times [K_1,K] \le L$.