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??

You need to fill in some important details, but the basic idea is fine. You might start like this:

Let $\langle x_n:n\in\Bbb N\rangle$ be a Cauchy sequence in $\langle X,e\rangle$. For $n\in\Bbb N$ let $t_n=\frac1{x_n}$. Then by definition for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $|t_m-t_n|<\epsilon$ whenever $n\ge n_\epsilon$. But that means that $\langle t_n:n\in\Bbb N\rangle$ is Cauchy in $\Bbb R$ with the usual metric, which is complete, so $\langle t_n:n\in\Bbb N\rangle$ converges to some $t\in\Bbb R$.

To finish, show that $t\ge 1$, let $x=\frac1t\in X$, and show that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ in $X$.
