Showing Euclidean metric and metric in $R^{2}$ produces same topology 
Question:
  Prove that the Euclidean metric and the metric $d_\infty \left ( x,y \right ):=\max\left \{ \left | x_{1}-y_{1} \right |,\left | x_{2}-y_{2} \right | \right \}$ defines the same topology in $\mathbb{R}^{2}$.

The Euclidean metric is defined as 
$d_2:\mathbb{R}^{n} \times \mathbb{R}^{n}\rightarrow \mathbb{R}$
$\vec{X} \times \vec{Y} \mapsto d\left ( \vec{X},\vec{Y} \right ) $
I would like to request for a useful hint to kickstart my attempt.
Thanks in advance.

To show that $d_1(x,y) = \sqrt{(x_1 - y_1)^2 + (x_2-y_2)^2}$ and the distance $d_2(x,y)=\max\{|x_1-$ $y_1|, |x_2-y_2|\}$ produce the same topology, it is enough to see the following inequality:

There exist constants $c_1$ and $c_2$ such that for all $x,y$, $c_1 d_2(x,y) \leq d_1(x,y) \leq c_2 d_2(x,y)$.

Proof: On one side, $$d_1(x,y) = \sqrt{(x_1 - y_1)^2 + (x_2-y_2)^2} \leq \sqrt{2\max((x_1 - y_1)^2,(x_2-y_2)^2)} \leq \sqrt{2}\cdot d_2(x,y)$$.
On the other hand, $$
d_2(x,y) = \max\{|x_1-y_1|, |x_2-y_2|\} = \sqrt{\max\{|x_1-y_1|, |x_2-y_2|\} ^2}
$$
Now,
$$
\sqrt{\max\{|x_1-y_1|, |x_2-y_2|\} ^2}\leq \sqrt{|x_1-y_1|^2  + |x_2-y_2| ^2} \leq d_1(x,y)
$$
Hence, the result follows.
From here, I'll leave you to show, using the definition for topology given by a metric, that the two topologies are the same. Hint: show that open sets given by one topology are contained in the other, and vice versa.
By the way, any two metrics derived from norms on $\mathbb{R}^2$ generate the same topology. You have only picked out a special case.
