Let $C_1 \subset \mathbb R^n$ and $C_2 \subset \mathbb R^m$ be convex sets. I want to prove that $\text{rint}(C_1\times C_2)=\text{rint}(C_1)\times \text{rint}(C_2)$.
I know that this question is already asked and answered, but I want to prove it by first proving that $\text{aff}(C_1\times C_2)=\text{aff}(C_1)\times \text{aff}(C_2)$ and then concluding that $\text{rint}(C_1\times C_2)=\text{rint}(C_1)\times \text{rint}(C_2)$.
By $\text{rint}$, I mean the relative interior of the set, and by $\text{aff}$, I mean the affine hull of the set.
Once you have that the affine hull commutes with the cartesian product, you can prove it like this:
"$\subset$": Let $x = (x_1, x_2) \in \text{ri}(C_1 \times C_2) \subset C_1 \times C_2$. Then there exists a $\varepsilon > 0$ such that $B_{\varepsilon}(x) \cap \text{aff}(C_1 \times C_2) \subset C_1 \times C_2$.
Since $B_{\frac{\varepsilon}{2}}(x_1) \times B_{\frac{\varepsilon}{2}}(x_2) \subset B_{\varepsilon}(x)$, we have \begin{align*} \left[ B_{\frac{\varepsilon}{2}}(x_1) \cap \text{aff}(C_1) \right] \times \left[ B_{\frac{\varepsilon}{2}}(x_2) \cap \text{aff}(C_2) \right] & = \left[ B_{\frac{\varepsilon}{2}}(x_1) \times B_{\frac{\varepsilon}{2}}(x_2) \right] \cap \left[\text{aff}(C_1) \times \text{aff}(C_2) \right] \\ & = \left[ B_{\frac{\varepsilon}{2}}(x_1) \times B_{\frac{\varepsilon}{2}}(x_2) \right] \cap \text{aff}(C_1 \times C_2) \\ & \subset B_{\varepsilon}(x) \cap \text{aff}(C_1 \times C_2) \subset C_1 \times C_2, \end{align*} so $B_{\frac{\varepsilon}{2}}(x_k) \cap \text{aff}(C_k) \subset C_k$ and thus $x_k \in \text{ri}(C_k)$ for $k \in \{ 1, 2 \}$.
"$\supset$": Let $x_k \in \text{ri}(C_k)$ for $k \in \{ 1, 2 \}$. Then there exist $\varepsilon_k > 0$ such that $B_{\varepsilon_k}(x_k) \cap \text{aff}(C_k) \subset C_k$. Let $x := (x_1, x_2) \in C_1 \times C_2$ and $\varepsilon := \min(\varepsilon_1, \varepsilon_2) > 0$. Due to $B_{\varepsilon}(x) \subset B_{\varepsilon}(x_1) \times B_{\varepsilon}(x_2)$ we have \begin{align*} B_{\varepsilon}(x) \cap \text{aff}(C_1 \times C_2) & = B_{\varepsilon}(x) \cap \left[\text{aff}(C_1) \times \text{aff}(C_2) \right] \\ & \subset \left[ B_{\varepsilon}(x_1) \times B_{\varepsilon}(x_2) \right] \cap \left[\text{aff}(C_1) \times \text{aff}(C_2) \right] \\ & = \left[ B_{\frac{\varepsilon}{2}}(x_1) \cap \text{aff}(C_1) \right] \times \left[ B_{\frac{\varepsilon}{2}}(x_2) \cap \text{aff}(C_2) \right] \\ & \subset C_1 \times C_2, \end{align*} so $x \in \text{ri}(C_1 \times C_2)$.