I'm reading a proof that says that the topologies on $\mathbb{R}^n$ induced by the euclidean metric $d$ and the square metric $p$ are the same as the product topology on $\mathbb{R}^n$. It goes like this:
Let $x = (x_1,\cdots, x_n)$ and $y=(y_1,\cdots, y_n)$ be two points of $\mathbb{R}^n$. Them:
$$p(x,y)\le d(x,y)\le \sqrt{n}p(x,y)$$
The first inequalioty shows that
$$B_d(x,\epsilon)\subset B_p(x,\epsilon)$$ since if $d(x,y)< \epsilon$ then $p(x,y)< \epsilon$
then the proof continues...
Well, why the inclusion of the balls is in that direction? For me, since $p(x,y)\le d(x,y)$, then the ball with respect to $p$ should be included in theone with respect to $d$
From the geometry, it is sort of obvious that $B_{d}\left(x, \epsilon\right) \subset B_{p}\left(x, \epsilon\right)$
However, you asked for a proof. First let us show $y \in B_{d}\left(x, \epsilon\right) \Rightarrow y \in B_{p}\left(x, \epsilon\right)$:
$$ y \in B_{d}\left(x, \epsilon\right) \\ \Leftrightarrow d(x, y) < \epsilon \\ \Rightarrow p(x, y) < \epsilon \\ \Leftrightarrow y \in B_{p}\left(x, \epsilon\right) $$
where the third line is the inequality $p(x, \epsilon) \le d(x, \epsilon)$. To show that $y \in B_{p}\left(x, \epsilon\right) \nRightarrow y \in B_{p}\left(x, \epsilon\right)$, consider the point $y=\left( x_1 + \alpha\epsilon, x_2 + \alpha \epsilon, \dots \right)^T$, then $p(x, y)=\alpha\epsilon$, while $d(x, y)=\sqrt{n}\alpha\epsilon$, so for $\frac{1}{\sqrt{n}} < \alpha < 1$, $y \not\in B_d\left(x, \epsilon\right)$.