Any nonempty open set in $\mathbb R^n$ contains a compact cube $C$ of volume $v:=|C|>0$. (All our cubes will be assumed to have edges parallel to the coordinate axes.)  

So, it is enough to show that $C$ cannot be covered by a set $S$ of (say) open cubes with total volume $<v$. By compactness, without loss of generality the cardinality (say $k$) of the set $S$ is finite. 

Replace now the compact cube $C$ by the corresponding left-open cube contained in $C$ of the same volume $v$, and still denote this left-open cube by $C$. A left-open cube here means the Cartesian product of left-open intervals. 
Also, replace each open cube $c\in S$ by the corresponding left-open cube containing $c$ of the same volume $|c|$, and still denote the resulting (covering $C$) finite set of left-open cubes by $S$. Moreover, by a [semiring ][1] argument, without loss of generality the left-open cubes $c\in S$ are pairwise disjoint and their union is $C$. 

Take any (left-open) cube $c\in S$. If all the hyperplanes through all the faces of $c$ do not intersect the interior of $C$, then either $c$ contains $C$ or is disjoint with $C$. So, if there is no $c\in S$ such that at least one of the hyperplanes through all the faces of $c$ intersects the interior of $C$, then at least one $c\in S$ contains $C$, and we are done. 

Otherwise, there is some $c_*\in S$ such that the hyperplane $H$ through some one of the faces of $c_*$ intersects the interior of $C$. This hyperplane $H$ cuts $C$ into two disjoint left-open cubes, say $C_1$ and $C_2$. Also, $H$ cuts each $c\in S$ into two disjoint left-open cubes. Moreover, for each $j\in\{1,2\}$, the number of nonempty left-open cubes of the form $C_j\cap c$ for $c\in S$ is $<k$. Therefore and because $|C|=|C_1|+|C_2|$, the desired result follows by induction on $k$. 


  [1]: https://en.wikipedia.org/wiki/Ring_of_sets#Related_structures