Is this proof involving complete metric spaces correct?

Question

Show that if every closed ball of a metric space $(X, d)$ is complete then $ X$ is complete.

I thought the following: given $(x_n)$ a Cauchy sequence in $X$, we have that the set $A= \{x_{1}, x_{2},..., x_{n},...\} $ is bounded, that is, $ diam(A) < M $ for some $ M \in \mathbb {R}$ Therefore for any $ n_{0}\in \mathbb {N}$, $ A \subset B [x_{n_{0}}, M]$ From the hypothesis we know that every closed ball is complete and therefore $ x_{n} \rightarrow x \in B [x_{n_{0}}, M] \subset X $ And so $ X $ is complete.

For some reason I am not completely sure my proof is correct, could anyone tell me what you think?

Answer 1

The proof is correct.

As an $\epsilon$-improvement for clarity, I would change the last sentence with something like:

$(x_n)$ Cauchy, $B[x_{n_0},M]$ complete and $(x_n)\subset B[x_{n_0},M]$ imply that $(x_n)$ converges to some $x\in B[x_{n_0},M]\subset X$. It follows that $X$ is complete.