[proofplan] We prove the result directly from the definitions. Given an arbitrary tolerance $\varepsilon>0$, convergence to $x$ gives a common index $N$ after which every term $x_n$ lies within $\varepsilon/2$ of $x$. Then any two terms $x_m$ and $x_n$ with $m,n\ge N$ are close to each other because the metric triangle inequality bounds their distance by the sum of their distances to the common limit $x$. [/proofplan] [step:Use convergence with tolerance $\varepsilon/2$] Let $\varepsilon>0$ be arbitrary. Since $x_n\to x$ in the metric space $(X,d)$, applying the definition of convergence with tolerance $\varepsilon/2>0$ gives an index $N\in\mathbb{N}$ such that, for every $k\in\mathbb{N}$ with $k\ge N$, \begin{align*} d(x_k,x)<\frac{\varepsilon}{2}. \end{align*} [guided] We must prove the Cauchy condition, so the final estimate must compare two far-out terms $x_m$ and $x_n$. The convergence hypothesis compares each far-out term to the limit point $x$. Therefore we choose the tolerance $\varepsilon/2$ so that two such comparisons can be added and still remain below $\varepsilon$. Let $\varepsilon>0$ be arbitrary. Since $x_n\to x$ in $(X,d)$, the definition of convergence says that for every positive tolerance there is an index after which all terms are within that tolerance of $x$. We apply this definition to the positive number $\varepsilon/2$. Hence there exists $N\in\mathbb{N}$ such that, for every $k\in\mathbb{N}$ with $k\ge N$, \begin{align*} d(x_k,x)<\frac{\varepsilon}{2}. \end{align*} This single index $N$ will be used simultaneously for both indices in the Cauchy condition. [/guided] [/step] [step:Compare two tail terms through the limit point] Let $m,n\in\mathbb{N}$ satisfy $m\ge N$ and $n\ge N$. By the triangle inequality for the metric $d$, applied to the three points $x_m,x,x_n\in X$, \begin{align*} d(x_m,x_n)\le d(x_m,x)+d(x,x_n). \end{align*} Because $d$ is symmetric, $d(x,x_n)=d(x_n,x)$. The choice of $N$ gives \begin{align*} d(x_m,x)<\frac{\varepsilon}{2} \end{align*} and \begin{align*} d(x_n,x)<\frac{\varepsilon}{2}. \end{align*} Therefore \begin{align*} d(x_m,x_n)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \end{align*} [/step] [step:Conclude the Cauchy condition] We have shown that for every $\varepsilon>0$ there exists $N\in\mathbb{N}$ such that for all $m,n\in\mathbb{N}$ with $m,n\ge N$, \begin{align*} d(x_m,x_n)<\varepsilon. \end{align*} This is exactly the definition of a Cauchy sequence in the metric space $(X,d)$. Hence $(x_n)_{n=1}^{\infty}$ is Cauchy. [/step]