Fix $\varepsilon > 0$. Since $x$ is convergent in $M$, then result
R1089: Characterisation of convergent sequences in metric space shows that there is a point $a \in X$ and a threshold $N \in \mathbb{N}$ such that $d(x_n, a) < \varepsilon / 2$ for every $n \geq N$. Thus, if $n_0, n_1 \in \mathbb{N}$ such that $n_0, n_1 \geq N$, then
\begin{equation}
d(x_{n_0}, x_{n_1})
\leq d(x_{n_0}, a) + d(a, x_{n_1})
< \frac{\varepsilon}{2} + \frac{\varepsilon}{2}
= \varepsilon
\end{equation}
Since $\varepsilon > 0$ was arbitrary, the proof is complete. $\square$
