Prove that $\mathbb{Q}^n$ is dense in $\mathbb{R}^n$ w.r.t these metrics:
$d_{2}(x,y)=(\sum_{k=1}^{n}(x_{k}-y_{k})^2)^\frac{1}{2}$
$d_{0}(x,y))=\max_{1\leq k\leq n}|x_{k}-y_{k}|$
my attempt of proving
$\mathbb{Q}^n$ is dense in $\mathbb{R}^n\Leftrightarrow $ $\mathbb{R}^n$ is contained in closure of $\mathbb{Q}^n$
So take $x\in \mathbb{R}^n$ and try to prove $x\in closure\; of \; \mathbb{Q}^n$
I know the idea but i can't write a rigorous proof
Hints:
$(1)$ $d_0$ and $d_2$ give rise to same topology on $\mathbb{R}$. This topology is the usual topology on $ \mathbb{R} $. In the case of $n=2$, a ball of radius $1$ in $d_0$ looks like a disc, and in $d_2$ looks like a square. Intuitively, inside each disc we can find a square and inside each square a disc. Extend similar arguments to the general case.
$(2)$ A dense set has a nonempty intersection with every nonempty open subset of the space.
$(3)$ Show that $\mathbb{Q}$ is dense in $\mathbb{R}$. This is not so difficult if you use Hint $(2)$.Using this, show that $\mathbb{Q^n}$ is dense in $\mathbb{R^n}$. For a point $x=(x_1,x_2,...x_n) \in \mathbb{R^n}$ construct a sequence that converges to it. This can be done by thinking about constructing a sequence for each individual $x_i$.