Prove that Euclidean distance and Uniform norm generate the same topology I'm teaching myself topology using a book I found. I need help proving the following.

Suppose $M$ is the unit disk in $\mathbb{R}^2$, $d$ is the Euclidean distance and $d'$ is the Uniform norm. Prove that $d$ and $d'$ generate the same topology on $M$.

I want to do this using the following definition of topology.

 is a topology on X if and only if the following are true:
  (i) X and Ø are elements of .
  (ii)  is closed under finite intersections.
  (iii)  is closed under arbitrary unions.

and $without$ using:

$d$ and $d'$ generate the same topology on $M$ if and only if for every $x \in M$ and every $r>0$, there exist positive numbers $r_1$ and $r_2$ such that $B_{r_1}^{(d')}(x) \subseteq B_{r}^{(d)}(x)$ and $B_{r_2}^{(d)}(x) \subseteq B_{r}^{(d')}(x)$.

I eventually want to prove the above theorem, but I want to start with a specific example first. 
In finite sets I showed that two typologies were the same by listing all the subsets, and their unions and intersections. However there is no way to list them all because $M$ is an Uncountable set.
I was considering an RAA proof starting with the assumption that there does exist a subset of one topology that does not exist in the other, but then I don't know where to go from there. I was also considering an algorithm that puts circles in a square lattice, and then decreases the radii of each circle while increasing the amount of circles in the square. Talking this process to infinity, the circles can fill the square. 
In short, I have no idea how to even being proving this. 

In this case we have two inequalities:
$(x-a)^2 + (y-b)^2 \le 2\max((x-a)^2, (y-b)^2) = 2(\max(|x-a|,|y-b|))^2$, and taking square roots we have $d_2((x,y),(a,b)) \le \sqrt{2}d_\infty((x,y),(a,b))$, where $d_2$ denotes the Euclidean distance and $d_\infty$ uniform norm.
$|x-a|^2 = (x-a)^2 \le (x-a)^2 + (y-b)^2$, so $|x-a| \le d_2((x,y),(a,b))$, and similarly $|y-b| \le d_2((x,y),(a,b))$, so also for their maximum, and hence $d_\infty((x,y)(a,b)) \le d_2((x,y),(a,b))$.
This will allow you to define the $r_1$ and $r_2$.
