My confusion I think stems from lack of understanding of Borel sets. I saw this exact question basically but I didn't understand the portion that's confusing me. Let me state the relevant part of the proof briefly:
Let $E\in\mathcal{B}(\mathbb{R}^n)$, $F\in\mathcal{B}(\mathbb{R}^m)$. Then $E\times F \in\mathcal{B}(\mathbb{R}^{n+m})$.
Step 1: Pick an open set $D\subset \mathbb{R}^n$ and consider $\{F\subset \mathbb{R}^m:D\times F\in\mathcal{B}(\mathbb{R}^{n+m})\}$. This is a $\sigma$-algebra.
Is there a certain way to go about showing that $D^c\times F$ is also a Borel set, or should that just be obvious?
$\newcommand{\BB}{\mathcal{B}}\newcommand{\RR}{\mathbb{R}}\newcommand{\AA}{\mathcal{A}}$ So you want to show that $\AA_D=\{F\subset\RR^m : D\times F \in\BB(\RR^{n+m})\}$ is a $\sigma$-algebra. So if $F\in\AA_D$, you want to show $F^c\in\AA_D$. So we have that $D\times F$ is Borel, and $D\times \RR^m$ is open, and hence Borel.
Now we have that $$(D\times F)^c \cap (D\times \RR^m)= (D\times \RR^m)\setminus (D\times F) = D\times (\RR^m \setminus F) = D\times F^c.$$ Therefore $D\times F^c$ is Borel, so $F^c \in \AA_D$.