$\vect{I}$ is a generalized rectangle in $\R^n$ with edges of length
$l_i$. $\vect{J} \subseteq
\vect{I}$ is the main character here. If the edges of $\vect{J}$ is
allowed to coincide with those of $\vect{I}$, then the problem is
trivial.
How about finding $\vect{J} \subseteq \interior \vect{I}$ such that
$\volume \vect{I} - \volume \vect{J} < \varepsilon$?
Goal: Choose a big $L$ that depends on $\vect{I}$ and
$\varepsilon$.
\[
\begin{split}
\volume \vect{J}
:=& \prod_{i = 1}^n \left(l_i - \frac{\varepsilon}{L}\right) \\
\volume \vect{I} =& \prod_{i = 1}^n l_i \\
=& \prod_{i = 1}^n \left [\left(l_i - \frac{\varepsilon}{L}\right) +
\frac{\varepsilon}{L} \right] \\
=& \left[ \prod_{i = 1}^n \left( l_i - \frac{\varepsilon}{L} \right)
\right] + \sum_{i = 1}^n \left( l_i -
\frac{\varepsilon}{L} \right) \left( \frac{\varepsilon}{L}
\right)^{n - 1} \\
&+ \sum_{i = 2}^{n - 1} \sum_{\left\{ a_j \right\} \in S_i}
\left[ \prod_{j = 1}^i \left( l_{a_j}
- \frac{\varepsilon}{L} \right) \right]
\left( \frac{\varepsilon}{L} \right)^{n - i}
+ \left( \frac{\varepsilon}{L} \right)^n,
\end{split}
\]
where $S_i$ is a set of $i$ numbers chosen from $\left\{ 1,\dots,n
\right\}$, and $\left\{ a_j \right\}$ is an increasing sequence in
$S_i$.
I assumed that $n > 2$. For $n = 2$, deleting the third term in the
last step will do. For $n = 1$, the problem is trivial.
Then
\[
\begin{split}
\volume \vect{I}
=& \volume \vect{J} + \sum_{i = 1}^{n - 1}
\sum_{\left\{a_j\right\} \in S_i}
\left[ \prod_{j = 1}^i \left( l_{a_j}
- \frac{\varepsilon}{L} \right) \right]
\left( \frac{\varepsilon}{L} \right)^{n - i}
+ \left( \frac{\varepsilon}{L} \right)^n \\
<& \volume \vect{J} + \sum_{i = 1}^{n - 1}
\sum_{\left\{ a_j \right\} \in S_i}
\left( \prod_{j = 1}^n l_{a_j} \right)
\left( \frac{\varepsilon}{L} \right)^{n - i}
+ \left( \frac{\varepsilon}{L} \right)^n \\
=& \volume \vect{J} + \sum_{i = 1}^{n - 1}
\sum_{\left\{ a_j \right\} \in S_i}
\volume \vect{I} \, \left( \frac{\varepsilon}{L} \right)^{n - i}
+ \left( \frac{\varepsilon}{L} \right)^n \\
=& \volume \vect{J} + \sum_{i = 1}^{n - 1} {n \choose i}
\volume \vect{I}\,\left( \frac{\varepsilon}{L} \right)^{n - i}
+ \left( \frac{\varepsilon}{L} \right)^n
\end{split}
\]
Rearanging the above inequality gives
\[
\volume \vect{I} - \volume \vect{J} < \sum_{i = 1}^{n - 1} {n \choose
i} \volume \vect{I} \, \left( \frac{\varepsilon}{L} \right)^{n - i} +
\left( \frac{\varepsilon}{L} \right)^n
\]
Assume $\varepsilon < 1, L \ge 1$, and let $M := \max\limits_{1 \le i
\le n - 1} {n \choose i}$.
Then we have
\[
\begin{split}
\volume \vect{I} - \volume \vect{J}
<& \left( \frac{\varepsilon}{L} \right)^n + \sum_{i = 1}^{n - 1} {n
\choose i} \volume \vect{I} \, \left( \frac{\varepsilon}{L}
\right)^{n - i} \\
=& \frac{\varepsilon}{L} + M (n - 1) \volume \vect{I}
\,\frac{\varepsilon}{L} \\
=& [1 + M (n - 1) \volume \vect{I}] \,\frac{\varepsilon}{L}
\end{split}
\]
Hence, if $L \ge 1 + M (n - 1) \volume \vect{I}$, then
$\volume \vect{I} - \volume \vect{J} < \varepsilon$.