proof that metrics generate the same topology, if their balls can be contained in one another. I'm teaching myself topology using a book I found. One of the exercises are to prove the following:

Prove, $d$ 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 took me a long time to comprehend the concepts, but my actual proof is very brief. I just want to know if my proof is valid or not. 
My proof involves the definition of a convergent sequence:

A sequence, $q_n$, of elements of a metric space $(X, d)$ is said to converge to $p \in X$, ($q_n \rightarrow p$) if for any positive epsilon an integer $N_\varepsilon$ such that $d(q_n, p) < \varepsilon$ for all $n \geq N_\varepsilon$. 

And the following theorem:

Two topologies, generated by a metrics d and d’, are equivalent iff the set of d-convergent sequences coincides with the set of d′-convergent sequences.

Now the actual proof:
A $d$-convergent sequence must converge to a point, $p$. By the definition of $d$-convergent sequence, all $q_n$, after $n \geq N_\varepsilon$, will be contained in an open $d$-ball, of radius $r$. If it is possible to make a $d$ and $d’$-balls, around $p$, contained within one another, then the $d$-convergent sequence must also be a $d’$-convergent sequence, as shown in the image below.

(above) the green circles are $d$-balls, and blue square is a $d'$-ball. The sequence is contained in the inner green circle, and that inner circle is contained in the square. Thus the sequence is also contained in the square. 
The converse is also true because if you make another $d'$-ball in the inner $d$-ball, the same argument can be made. 
QED
 A: Your approach is somewhat roundabout. Just use the fact that open balls generate the topology.
So assume $d$ and $d'$ generate the same topology.
Having $x \in M$, and $r>0$, we know that $B_r^{(d)}(x)$ is by definition an open set in the topology generated by $d$. So it is open in the topology generated by $d'$, so it must be a union of $d'$-open balls. In particular $x \in B_r^{(d)}(x)$ must be in such a ball, so there exists some $r_1>0$ such that $x \in B_{r_1}^{(d')}(x) \subseteq B_r^{(d)}(x)$. The same can be said for the open set $B_r^{(d')}(x)$ which is in the topology generated by $d'$ so in the one generated by $x$ etc., so we can find $r_2$ as stated.
Now suppose we have this condition on the balls. Now let $O$ be open in the $d$-topology. We want to show it is open in $d'$-topology, so pick $x \in O$. $O$ is open in the $d$-topology so there is some $r>0$ such that $x \in B_r^{(d)}(x) \subseteq O$ (open sets are unions of balls...). Now apply the condition and find $r_1 > 0$ such that $(x \in ) B_{r_1}^{(d')}(x) \subseteq B_r^{(d)}(x) (\subseteq O)$, which shows that $x$ has a $d'$-ball that sits inside $O$, and as we can do this for all $x \in O$, $O$ is $d'$-open. 
That a $d'$-open $O$ is also $d$-open can be done using the other half of the condition (the $r_2$ one), check this similarly. 
