Let $R$ be the closed rectangle $[a_1, b_1] \times \ldots \times [a_n, b_n]$. I am trying to show that if $R \subseteq (x + S)$, where $x \in \mathbb{R}^n$ and $S$ is a subspace of $\mathbb{R}^n$ of dimension $m < n$, then the volume of $R$, $(b_1 - a_1) \times \ldots \times (b_n - a_n)$, must equal 0. Here's what I currently have:
Suppose that $S$ has a basis consisting of standard basis vectors (vectors with one nonzero coordinate that is equal to 1). Without loss of generality, assume that $\{e_1, \ldots, e_m\}$ is a basis of $S$. Then, for each $y \in R$, $y = x + c_1 \cdot e_1 + \ldots + c_m \cdot e_m$, where $c_1, \ldots, c_m \in \mathbb{R}$. Coordinate $m + 1$ of $y$ equals coordinate $m + 1$ of the expression on the right-hand side, i.e., $y_{m + 1} = x_{m + 1}$. It follows that the $(m + 1)$st coordinate of each point in $R$ is $x_{m + 1}$. In particular, the $(m + 1)$st coordinates of $(a_1, \ldots, a_n)$ and $(b_1, \ldots, b_n)$ equal $x_{m + 1}$: $a_{m + 1} = x_{m + 1} = b_{m + 1}$. Thus, $b_{m + 1} - a_{m + 1} = 0$, so the volume of $R$ is 0.
Based on some examples in $\mathbb{R}^2$ and $\mathbb{R}^3$ that I considered, it seems like $S$ should always have a basis consisting of standard basis vectors, possibly because the sides of $R$ are parallel to the axes. I haven't been able to prove this; is this true, and if so, how can it be proven?
No that is not true in general. Here's a counter example. Let $R=[0,1]\times[0,0]\times[0,0]$, $S=\left\{\lambda(1,0,0)+\mu(1,1,1) \ \mid \ \lambda,\mu \in \mathbb{R}\right\} \subset \mathbb{R}^3$. We clearly have that $R\subset 0+S $ and $S$ is two dimensional but no two of the standard basis vectors span it.
Here's how you can prove your statement without that assumption. Let $e_j$ are the standard basis vectors of $\mathbb{R}^n$. Let's set $$h_i := b_i - a_i. $$ $$a = (a_1, \dots,a_n) $$ Now we have that $$ R = \left\{ a + \sum_{i=0}^n \lambda_ih_ie_i \ \mid \ 0 \leq \lambda_i \leq 1 \right\} $$ Since $ R \subset x + S$ we know that the difference of two vectors of $R$ will be in $S$ because if $x + s\in R$ and $x+ s^\prime \in R $ then $ x + s - (x+ s^\prime) = s-s^\prime \in S$. But now by considering the vectors $a + h_ie_i$ and $a$ which are in $R$ by subtracting them we can see that their difference $h_ie_i$ is in $S$. Since vector space is closed under linear combination their span is contained in $S$ as well. However since $S$ is less than $n$ dimensional we must have that some $h_i=0$.