Let $(X,d)$ be a metric space. Prove that the function $$\rho(x,y) = \frac{d(x,y)}{ 1 + d(x,y)}$$ defines an equivalent metric on $X$. (Metrics $d_1$ and $d_2$ are called equivalent, if every convergent sequence in the metric $d_1$ converges also to the same point in the metric $d_2$ and vice-versa).
Attempt: Assume there is a convergent sequence $x_n$ that converges to $x$ in metric $d(x,y)$. Thus, given $\epsilon>0$ $\exists N$ such that $\forall n > N$ we have $0 \leq d(x_n,x)< \epsilon$. Then, \begin{align*} 1 \leq 1+ d(x_n,x) < 1+ \epsilon \\ \implies \frac{1}{1+ d(x_n,x)} \leq 1 \\ \implies 0 \leq \frac{d(x_n,x)}{1+ d(x_n,x)} < \frac{\epsilon}{1+ d(x_n,x)} < \epsilon \\ \implies \rho(x_n,x) < \epsilon\\ \end{align*}
Thus $x_n$ that converges to $x$ in metric $\rho(x,y)$.
Now, assume $x_n$ that converges to $x$ in metric $\rho(x,y)$. Thus, given $\epsilon>0$ $\exists N$ such that $\forall n > N$ we have $0 \leq \rho(x_n,x)< \epsilon$. \begin{align*} \frac{d(x_n,x)}{1+ d(x_n,x)} < \epsilon \\ d(x_n,x) < \epsilon + \epsilon d(x_n,x)\\ d(x_n,x) < \frac{\epsilon}{1-\epsilon} \\ \end{align*}
Is the procedure correct? Also I am stuck at the end. I cannot show $d(x_n,x) < \epsilon$
Please note: There are solutions in this website that show equivalence but use a different definition for the equivalent metric.
Edit: (Based on Arthur's suggestion)
assume $x_n$ that converges to $x$ in metric $\rho(x,y)$. Thus, there $\exists N$ such that $\forall n > N$ we have $0 \leq \rho(x_n,x)< \frac{\epsilon}{1+\epsilon}$. \begin{align*} \frac{d(x_n,x)}{1+ d(x_n,x)} (1+ \epsilon) < \epsilon \\ d(x_n,x) (1+\epsilon)< \epsilon + \epsilon d(x_n,x)\\ d(x_n,x) < \epsilon \\ \end{align*}
Thus $x_n$ converges to $x$ in metric $d(x,y)$.
Let $$ \rho = \frac {d}{1+d}$$
Note that for $$0\le \rho \le 2/3$$ we have $$ \rho \le d \le 3\rho $$
Thus $$ d\to 0 \iff \rho \to 0$$
The convergence in one metric is equivalent to the convergence to the other metric.
Thus the metrics are equivalent.