Characterisation of equivalent metrics 
Definition (equivalent metrics)
Two metrics on a set $X$ are called equivalent if for all $x \in X$
$$
\forall \varepsilon > 0 \ 
\exists r > 0:
U_1(x,r) \subset U_2(x,\varepsilon)
\quad\text{and} \quad
U_2(x,r) \subset U_1(x,\varepsilon),
$$
where $U_i(x,r) := \{ y \in X: d_i(x,y) < r\}$ for $i \in \{1,2\}$.



Task: Show that two metrics $d_1$ and $d_2$ are equivalent if and only if $d_1$ and $d_2$- convergent sequences coincide.

Is there a constructive proof for the direction of "$\impliedby$"?
This is my approach: "$\implies$" should be correct and hopefully my approach with contraction for "$\impliedby$", too,  but feel free to correct any mistakes, might they be mathematically or about proof-writing.
Proof.
"$\implies \DeclareMathOperator{\N}{\mathbb{N}} \DeclareMathOperator{\eps}{\varepsilon}$":
Let $x \in X$ and $\eps > 0$ and $(x_n)_{n \in \N} \subset X$ $d_1$-convergent to $x$.
This means there exits an $N_{\eps} \in \N$ such that
\begin{equation*}
\{ x_n: n \ge N_{\eps} \} \subset  U_1(x,r) \subset U_2(x, \eps),
\end{equation*}
where the second inclusion and the existence of such an $r > 0$ is due to $d_1$ and $d_2$ being equivalent.
"$\impliedby$":
Let $x \in X$ and $\eps>0$ and define $U^{(n)} := U_1\left(x,\frac{1}{n}\right)$.
Any sequence with $x_n\in U^{(n)}$ for all $n \in \N$ converges to $x$ in $(X,d_1)$.
Towards contradiction assume that there exists no $r > 0$ such that $U_1(x,r) \subset U_2(x,\eps)$ holds.
Then we can find a sequence with $x_n\in U^{(n)}$ with $x_n\notin U_2(x,\eps)$ but that would imply that $x_n$ is not convergent to $x$ in $(X,d_2)$, a contradiction.
Therefore, we find a $r_1>0$ satisfying $U_1(x,r_1)\subset U_2(x,\eps)$.
For reasons of symmetry we can find a $r_2$ satisfying $U_2(x,r_2)\subset U_1(x,\eps)$. Choosing $r=\min\{r_1,r_2\}$ finishes the proof.
Remark: I know that a equivalent charaterisation of equivalent metrics is that for all $x,y \in X$ there exists constants $C >0$ and $c := C^{-1} > 0$ such that $$c d_1(x,y) \le d_2(x,y) \le C d_1(x,y).$$ The equivalence between this characterisation and the convergence property is clear, and not what I am trying to prove.
 A: $\impliedby$ First of all the following remark is useful:
The metrics $d_1$ and $d_2$ induce two topologies $\tau_1$ and $\tau_2$ that are equal if and only if $d_1$ and $d_2$ are equivalent.
$\rightarrow$ 
Let $x\in X$ and $\epsilon>0$. You can consider the ball $B^{d_1}(x,\epsilon)$ that is an open set on $\tau_1=\tau_2$ so, by definition of Topology induced by a metric, it is a union of some balls with respect to $\tau_2$:
$B^{d_1}(x,\epsilon)= \bigcup_{y\in A, s\in I} B^{d_2}(y, s)$ for an appropriate subset $A\subset X$ and $I\subset \mathbb{R}^+_0$
In particular there exists $z\in X$ and $S>0$ such that 
$B^{d_2}(z, S)\subseteq B^{d_1}(x,\epsilon)$ and $x\in 
B^{d_2}(z, S)$
Let $r_1:=S-d_2(x,z)>0$, then 
$B^{d_2}(x, r_1)\subseteq B^{d_2}(z, S)\subseteq B^{d_1}(x,\epsilon)$ 
So $B^{d_2}(x,r_1)\subseteq B^{d_1}(x,\epsilon)$
Now we can do the same reasoning considering te ball $B^{d_2}(x,\epsilon)$, that is an open set of the Topology $\tau_2=\tau_1$, so, by definition of Topology induced by a metric, we have 
$B^{d_2}(x,\epsilon)= \bigcup_{y\in B, s\in J} B^{d_1}(y, s)$ for an appropriate subset $B\subset X$ and $J\subset \mathbb{R}^+_0$
In particular there exists $z\in X$ and $S>0$ such that 
$B^{d_1}(z, S)\subseteq B^{d_2}(x,\epsilon)$ and $x\in 
B^{d_1}(z, S)$
Let $r_1=S-d_1(x,z)>0$, then 
$B^{d_1}(x, r_2)\subseteq B^{d_1}(z, S)\subseteq B^{d_2}(x,\epsilon)$ 
so
$B^{d_1}(x,r_2)\subseteq B^{d_2}(x,\epsilon)$
To sum up we have 
$B^{d_2}(x,r_1)\subseteq B^{d_1}(x,\epsilon)$ and
$B^{d_1}(x,r_2)\subseteq B^{d_2}(x,\epsilon)$
If we choose $r:=\min{r_1,r_2}>0$, then we have 
$B^{d_2}(x,r)\subseteq B^{d_1}(x,\epsilon)$ and
$B^{d_1}(x,r)\subseteq B^{d_2}(x,\epsilon)$
This is exactly the condition of equivalence  between two metrics. 
$\leftarrow$
We suppose that two metrics must be equivalent and we want to prove that the topologies induced by that metrics are equal.
We know that the set of all balls of a metric space is a base for the topology induced by the metric, so we must only prove that each ball of $(X,d_1)$ is an open set with respect to $\tau_2$ and viceversa. 
Let $B^{d_1}(x,\epsilon)$ be a ball of the metric space $(X,d_1)$. Let $y\in B^{d_1}(x,\epsilon)$ and $\epsilon_y:=\epsilon-d_1(x,y)$. Then it is clear that 
$B^{d_1}(y,\epsilon_y)\subseteq B^{d_1}(x,\epsilon)$
By definition of equivalence of two metrics  there exists $r_y>0$ such that 
$B^{d_2}(y,r_y)\subseteq B^{d_1}(y,\epsilon_y)$ 
and 
$B^{d_1}(y,r_y)\subseteq B^{d_2}(y,\epsilon_y)$
So for each $y\in B^{d_1}(x,\epsilon)$ there exists $r_y>0$ such that 
$B^{d_2}(y,r_y)\subseteq B^{d_1}(y,\epsilon_y)\subseteq B^{d_1}(x,\epsilon) $ 
This means that 
$ B^{d_1}(x,\epsilon)= \bigcup_{y\in B^{d_1}(x,\epsilon} B^{d_2}(y,r_y)\in \tau_2$ 
So we have that $\tau_1\subseteq \tau_2$. 
In a similar way you can prove that $\tau_2\subseteq \tau_1$. 
Now we return to our problem. 
We suppose that  $d1, d2$ convergent sequences coincide and we want to prove that the two metrics are equivalent. 
The previous remark tells us that we can simply prove that $\tau_1=\tau_2$. 
You can observe the following remark: 
Let $X$ be a  first countable space and $A\subseteq X$ a subset of $X$. If $a\in X$ is a limit point for your set $A$, then there exists a sequence $\{a_n\}_n\subseteq A$ such that $a_n\to a$. This result is not true in general.
Proof: Let $\{U_n\}_n$ be a countable fundamental neighborhood system for the point $a$. For each $n\in \mathbb{N}$ you can define 
$U’_n:=\bigcap_{m\leq n}U_m$ 
In this case you have that $\{U’_n\}_n$ is again a countable fundamental neighborhood system for the point $a$. 
So we can choose for each $n\in \mathbb{N}$ a point $a_n\in U’_n\cap A\neq \emptyset$. Then $a_n\to a$ in fact if you consider an open set $B$ of $a$, there exists $U’_N\subseteq B$ so for each $n\geq N$ 
$a_n\in U’_n\subseteq U’_N\subseteq B$.
This result is not true in general because for example if you consider a non countable set $X$ with co-countable $\tau$ topology, then there is non trivial sequence that converges to some point $a$. In fact if by contradiction there is a non trivial sequence $\{a_n\}_n$ that converges to $a$, then the sequence $\{a_n\}_n$ is a countable set and the open set $X/\{a_n\}_n$ is an open set of the point $a$ that does not contains elements of the sequence and it is not possible. 
It is true in general that each metric space is first countable because for each point $a$ 
$\{B(a,\frac{1}{n})\}_n$ is a countable fundamental neighborhood system. 
Let $C$ be a closed set of $(X,d_1)$
and let $a$ be a limit point for $C$ with respect to the space $(X,d_2)$. 
By the previous remark, there exists a sequence $\{a_n\}_n\subseteq C$ convergent to $a$ in $(X,d_2)$. By hypothesis the $d_1-d_2$ convergent sequence coincide, so $a_n\to a$  in $(X,d_1)$ but $C$ is closed in this space, so $a\in C$. 
This means that 
$cl_{d_2}(C)=C$ so $C$ is a closed set in $(X,d_2)$. 
Hence we have $\tau_1\subseteq \tau_2$. 
In a similar way you can prove that if $C$ is a closed set in $(X,d_2)$, then it is also a closed set in $(X,d_1)$. 
This means that $\tau_2\subseteq \tau_1$. 
So $\tau_1=\tau_2$ that tells us the two metrics are equivalent.
