In a metric space, compactness implies completness

It contains some errors, though the basic idea is right. Here’s how I would prove that $\lim\limits_{n\to\infty}x_n=x^*$.

Let $\epsilon>0$. Since the sequence is Cauchy, there is an $n_0\in\Bbb N$ such that $d(x_m,x_n)<\frac{\epsilon}2$ for all $m,n\ge n_0$. Since $x^*$ is an accumulation point of the sequence, it is also an accumulation point of the tail sequence $\langle x_n:n\ge n_0\rangle$. (Why?) Thus, there is an $n_1\ge n_0$ such that $d(x_{n_1},x^*)<\frac{\epsilon}2$. The triangle inequality then implies that $d(x_n,x^*)\le d(x_n,x_{n_1})+d(x_{n_1},x^*)<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$ for all $n\ge n_0$. Since $\epsilon>0$ was arbitrary, $\langle x_n:n\in\Bbb N\rangle\to x^*$.