If a sequence $\{x_{n}\}$ is a Cauchy sequence and the sequence has a limit point $x_{0}$ then $x_{n} \rightarrow x_{0}$

We will prove that $\{x_n\}$ converges to that limit point $x_0$. Fix $\epsilon>0$ arbitrarily. Since $\{x_n\}$ is Cauchy, so there exists $N\in\mathbb{N}$ such that $d(x_n,x_m)\leq \frac{\epsilon}{2}$ for all $m,n\geq N$. Also as $x_0$ is a limit point of $\{x_n\}$, so there exists a subsequence $\{x_{n_k}\}_k$ such that $d(x_{n_k},x_0)\leq\frac{\epsilon}{2}$ for all $k\in\mathbb{N}$. Now choose some $n_k\geq N$ so that $d(x_{n_k},x_0)\leq\frac{\epsilon}{2}$ holds. Then $$d(x_n,x_0)\leq d(x_n,x_{n_k})+d(x_{n_k},x_0)\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\text{ for all }n\geq N.$$ And therefore $\{x_n\}$ converges to $x_0$.