The volume of a cover of an interval is greater than the volume of the interval 
Let $A \subset \mathbb{R}^n$ be a $n-$dimensional interval, i.e. $A = [a_1,b_1] \times [a_2,b_2] \times \cdots \times [a_n, b_n]$, where $[a_i, b_i]$ is an interval in $\mathbb{R}$, for every $i \in \{1,2,\cdots, n \}$. Prove that, for every cover $\{B_1, B_2, \cdots, B_m \}$ of $A$, where $B_i \subset \mathbb{R}^n$ is an $n-$dimensional interval for every $i \in \{1,2,\cdots,m \}$, we have that $$|A| \leq \sum_{i=1}^m |B_i|,$$ where $|I|$ denotes the volume of the interval $I$, i.e. if $I = [c_1, d_1] \times \cdots \times [c_n, d_n]$, we have that $$|I| = \prod_{i=1}^n (d_i - c_i). $$ 

It seems very trivial to me, but I do not know how to prove it using only the definition of a cover and algebraic manipulation.
I tried using the fact that if $\{B_1, \cdots, B_m \}$ is a cover of $A$, then, by letting $B_i = [a^{(i)}_1, b^{(i)}_1] \times \cdots \times [a^{(i)}_n, b^{(i)}_n]$, we have that $$\sum_{j=1}^m (b^{(j)}_i - a^{(j)}_i) \geq b_i - a_i, \forall i \in \{1,2,\cdots, n \},$$ but it does not give me the desired result.
 A: This question is typically an exercise when defining Riemann integration in $\mathbb{R}^n$, so I assume you're familiar with the concept of a partition of a rectangle etc. Otherwise, see Munkres' book Analysis on Manifolds or Spivak's Calculus on Manifolds. (They probably have a proof of this statement as well)
First, note that we can WLOG assume that each $B_i$ is contained in $A$ (why?). Having done this, construct a partition $P$ of the given rectangle $A$, by using the endpoints of the component intervals of the $B_i$'s. If you do this, then $A, B_1, \dots, B_m$ will all be a union of subrectangles of the partition $P$. This implies (try to justify yourself or refer to Munkres):
\begin{align}
|A| &= \sum_{S } |S| \\
|B_i| &= \sum_{S \subseteq B_i} |S| \qquad (1 \leq i \leq m),
\end{align}
where $\sum \limits_{S \subseteq B_i}$ means sum over all the subrectangles $S$ of the partition $P$ which are contained in $B_i$. Now the most crucial thing to notice is that every subrectangle $S$ of the partition $P$ is contained in some $B_i$. Or more precisely, for every $S$, there is an $i \in \{ 1, \dots, m\}$ such that $S \subseteq B_i$. Hence:
\begin{align}
|A| &= \sum_{S} |S| \\
&\leq \sum_{i=1}^m \sum_{S \subseteq B_i} |S| \\
&= \sum_{i=1}^m |B_i|
\end{align}
(The bold sentence tells us that every term in the first line already appears as a summand in the second line; which is why we have the $\leq$ sign).
