To show that $X = (0,1]$ is complete with respect to the metric $e $ where $e(x,y) = |\frac{1}{x} - \frac{1}{y}|$. Show that $X = (0,1]$ is complete with respect to the metric $e $ where  $e(x,y) = |\frac{1}{x} - \frac{1}{y}|$.
My proof: let $(x_n)$ be Cauchy in $(X,e)$. Let $(t_n) := \frac{1}{(x_n)}$. Then $(t_n)$ is Cauchy in $[1, \infty )$. Hence there exist a $t \in [1, \infty )$ such that $t_n \to t$ and this implies that $x_n \to 1/t$ in $(X,e)$. Thus....
Is the line of proof ok??

The idea is OK, but should be formalised a bit more.
Let $Y = [1,\infty)$ with the Euclidean metric $d$. Clearly, $Y$ is complete
as it is a closed subset of the complete $\mathbb{R}$ in the Euclidean metric.
Then $f(x) = \frac{1}{x}$ from $(X,e)$ to $(Y,d)$.
Then $d(f(x), f(y)) = |f(x) - f(y)| = |\frac{1}{x} - \frac{1}{y}| = e(x,y)$ for all $,y \in X$. So $f$ is an isometry. In particular, it is continuous, both ways, the same formula $g(x) = \frac{1}{x}$ defines an isometry from $(Y,d)$ to $(X,d)$.
So if $(x_n)$ is Cauchy, then $(t_n) = f(x_n)$ is Cauchy in $(Y,d)$, from being an isometry. And $(Y,d)$ is complete, so it has a limit $t$ and continuity of $g$ then garantuees $g(t_n) = (x_n)$ converges to $g(t)$ in $(X,e)$.
