Help in proving two metric are equivalent In attempting to prove that $d_1(x,y)=|x_1-y_1|+|x_2-y_2|$ and $d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$ induce the same (euclidean) topology on $\mathbb{R}^2$, I tried first to prove that any open set in the first metric space was open in the second, that is, for every point in a ball of $(\mathbb{R}^2, d_1)$, there existed a ball of $(\mathbb{R}^2, d_2)$ contained in it. 
After a lot of tries I couldn't find such a ball so I tried instead to prove that there must be such a ball. My reasoning (by contradiction) is as follows: 
Let $B_1(p,\epsilon)$ be an arbitrary ball in $(\mathbb{R}^2,d_1)$, and suppose this ball is not open in the euclidean topology. Then there must be some point $y$ pertaining to it such that its every euclidean neighbourhood is not contained in $B_1(p,\epsilon)$. Then, there must exist at least one $z \in B_2(y,\delta)$ (denoting the euclidean neighbourhoods of $y$) such that, for every $\delta>0$, $d_1(z,p)\geq \epsilon$.
But $$\epsilon \leq d_1(z,p)\leq d_1(z,y)+d_1(y,p)<2d_2(z,y)+d_1(y,p)<2\delta+\epsilon$$
Then, I reasoned that if $\delta$ can be arbitrariarly small, then $d_1(z,p)=\epsilon$. This implies that every neighborhoud lies on the boundary of the square, which is absurd given that $\delta$ can be any number. Is this ok? Can you give me some thoughts on how to improve, or even some other approach, like how to find the ball?
 A: Let $p = (x,y)$ be any point in $\mathbb{R}^2$ and let $r>0$.

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*Let $B_2 = B_2(p,r)$ be a ball in metric $d_2$. I'll show that $B_1 = B_1(p,r)\subset B_2$. Let $(x_0,y_0)$ be any point in $B_1$. Then
$$\sqrt{(x-x_0)^2+(y-y_0)^2} \leq |x-x_0|+|y-y_0|\leq r,$$
so $(x_0,y_0)\in B_2$. We found an open ball in $d_1$ that is contained in an open ball in $d_2$.

*Let $B_1 = B_1(p,r)$ be a ball in metric $d_1$. I'll show that $B_2 = B_2\left(p,\frac{\sqrt{2}}{2}r\right)\subset B_1.$
For that we need to know that
$$|a|+|b|\leq\sqrt{2a^2+2b^2}.$$
It follows from
$$(|a|-|b|)^2\geq 0\Leftrightarrow a^2+b^2\geq |a| |b|\Leftrightarrow 2a^2+2b^2\geq a^2+|a||b|+b^2=(|a|+|b|)^2.$$
Let $(x_0,y_0)$ be any point in $B_2$. Then
$$\begin{split}|x-x_0|+|y-y_0|&\leq \sqrt{2(x-x_0)^2+2(y-y_0)^2}=\sqrt{2}\cdot \sqrt{(x-x_0)^2+(y-y_0)^2}\leq\\
&\leq \sqrt{2}\cdot\frac{\sqrt{2}}{2}r=r,\end{split}$$
so $(x_0,y_0)\in B_1$. We found an open ball in $d_2$ that is contained in an open ball in $d_1$.
How did I know to pick $\frac{\sqrt{2}}{2}r$ as a radius for $B_2$? Easiest way is to draw these balls, and find a radius of a circle that is inscribed in a square with a diagonal of length $2r$. In the same way we see that there is no need to change a radius when we inscribe a square in a circle.
PS: This specific reasoning works when we look at the square 'generated' by metric $d_1$. There will be diffirent radiuses, when looking at a square 'generated' by Chebyshev metric (but still easy to find by drawing them).
