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. 
 A: (1). The def'n of the topology $T_d$ generated by a metric $d$ on a set $S$ is the topology whose base is the set of open $d$-balls. (I dk why you don't want to use the main theorem.)
(2). Two  metrics $d,d'$ on $S$ are inequivalent (meaning that $T_d\ne T_{d'}$) iff there exists $U\subset S$ which belongs to one of $T_d,T_{d'}$ but not the other.
Without loss of generality , suppose $U\in T_d$  \  $T_{d'}. $ Then, by (1), every $p\in U$ belongs to an open $d$-ball that is a subset of $U, $ but there must exist some $p\in U$ such that  no open $d'$-ball containing $p$ is a subset of $U.$ In particular $B_d'(p,1/n) \not \subset U$ for every $n\in \mathbb Z^+.$
So choose $q_n\in B_d'(p,1/n)$ \ $U.$ Then $\lim_{n\to \infty}d'(p,q_n)=0$ and $U\cap \{q_n:n\in \mathbb Z^+\}=\phi.$
On the other hand, for some $r>0$ we have $p\in  B_d(y,r)\subset U$ for some $y\in S$ and some $r\in \mathbb R^+. $ By the triangle inequality , $B_d(p,s)\subset B_d(y,r),$ where $s=\min( r-d(y,p), d(y,p)).$ Now $s>1/n$ for some $n\in \mathbb Z^+.$  So $d(q_n,p)\geq s$ for all but finitely many $n, $ and $\neg (\lim_{n\to \infty}d(q_n,p)=0).$
(3). Therefore if $d,d'$ are inequivalent ,there exists $p\in S$ and a sequence $(q_n)_n$ converging to $p$ in one metric but not the other.So if the set of $ d$-convergent sequences co-incides with the set of $d'$-convergent sequences, then $d,d'$ are equivalent.
(4).In your Q, show that $\lim_{n\to \infty}d((x_n,y_n),(x,y))=0$ and $\lim_{n\to \infty}d'((x_n,y_n),(x,y))=0$ are each equivalent to $[(x_n\to x)\land (y_n\to y)$ as $n\to \infty.] $ By (3), $d$ and $d'$ are equivalent.
(5)Remark: The converse of (3) is also true.  Two metrics are equivalent iff they have the same convergent sequences.
A: 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 $r_2$.
A: @Rolf Hoyer
quote: 

You want to show that any arbitrary open set in one topology will be open in the other.

I don't understand how that is sufficient. For example, I know that Euclidean metric and my "Clock" metric (see below) do not generate the same topology, on the set $[1,13)$.  However the open ball $B_{2}^{clock}(1)=(11,13) \cup [1,3)$, is open under the clock metric, and $(11,13) \cup [1,3)$ is open under the Euclidean metric on the set $[1,13)$; it is the union of two balls in Euclidean, but the union of 2 open sets in still open.  
.............................................................
"clock" metric, on the set [1,13) 
$$d(x,y)=6-||x-y|-6|$$
The distance between two numbers on a clock. Ex $d(2,11)=3$ (see diagram)


