Comparison between product measure and Lebesgue $(n+m)$-dimensional measure

Let's call $\mu_{n}$ the Lebesgue measure on $\mathbb R^n$. Prove that if $A \subset \mathbb R^n$ and $B\subset \mathbb R^m$ are open subsets then $$\mu_{n+m}(A\times B) = \mu_n(A)\mu_m(B).$$

In other words, this tells us that the Lebesgue $(n+m)$-dimensional measure and the product measure of $\mu_n$ and $\mu_m$ agree on open sets. How can I do? Suppose that $(A_i)_{i \in \mathbb N}$ and $(B_i)_{i \in \mathbb N}$ are sequences s.t. $$A \subseteq \bigcup_{i} A_i, \qquad B \subseteq \bigcup_{j} B_j$$ Then $$A \times B \subseteq \bigcup_{i} A_i \times\bigcup_{j} B_j$$ Now may I write - by $\sigma$-subadditivity $$\mu_{n+m}(A \times B) \le \sum_{i,j} \mu_n(A_i)\mu_m(B_j)$$ Is it correct? Now I would like to conclude - by taking $\inf$ on RHS - that $\mu_{n+m}(A \times B) \le \mu_n(A)\mu_m(B)$: how can I do? And what about the opposite inequality?

Bonus (self-posed) question: what happens if we remove the hypothesis $A,B$ are open? For example, is it true that if $\mu_n(A)<\infty$ and $\mu_m(B)<\infty$ then $$\mu_{n+m}^{\star}(A \times B) = \mu_n(A)\mu_m(B)$$ where $\mu^{\star}_{n+m}$ is the outer $(n+m)$-dimensional Lebesgue measure? Thanks in advance.

I suppose you want $(A_i)_i, (B_i)_i$ to be sequences of rectangular boxes. Taking the infimum over such sequences is ok and gives you "$\leq$".
But I think you prove the statement more easily with the uniqueness theorem for measures: For a rectangular box $R \in \mathbb{R}^m$ set $\alpha_R (A) = \mu_{n+m}(A \times R)$, $\beta_R (A) = \mu_n(A) \mu_m(R)$ (with the convention $0 \cdot \infty = 0$). Use the uniqueness theorem for measures to prove that $\alpha_R, \beta_R$ are coinciding measures on $\mathcal{B}(\mathbb{R}^n)$. Use it again to prove that $\mu_{n+m}(A \times \cdot)$ and $\mu_n(A) \mu_m(\cdot)$ coincide on $\mathcal{B}(\mathbb{R}^m)$ for all $A \in \mathcal{B}(\mathbb{R}^m)$.