Why is Q dense in R? Consider a topological space $S$ and an arbitrary subset $E$. Then if the closure of $E$ given by $\bar{E} = S$, we say that $E$ is dense in $S$.
How can we prove that $\mathbb{Q}^{n}$ is dense in the vector space $\mathbb{R}^{n}$ with the standard metric topology?
I intended to prove that $\mathbb{Q}^{n}$ is dense in the vector space $\mathbb{R}^{n}$ in the topological sense, not the usual "between two $\mathbb{R}$ there is a $\mathbb{Q}$". Of course, both might be equivalent, but that remains to be proved.

Something to think about (not a proof):
The closure of a subset is the minimal closed superset. What must a superset of $\mathbb{Q}$ look like?
It cannot have a maximal or minimal element, certainly.
Maybe you could make one out of closed intervals around each point of $\mathbb{Q}$, but it would be impossible to make each of this intervals disjoint (think about the space between each rational).
Trying to construct a closed superset always leads us to the whole real line, and hence $\mathbb{Q}$ is dense in $\mathbb{R}$. 
