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Prove that if every closed ball in a metric space X is a complete subspace, then X is complete.

The attempt at a solution.

Let ##\{x_n\}## be a cauchy sequence in X. Then, for ##ε=1##, ##\exists## ##n_0 \in \mathbb N##: ##\forall## ##m≥n_0##,##n≥n_0## ##d(x_m,x_n)<1##. Consider the closed ball centered at ##x_{n_0}## of radius 1. Then, the subsequence ##\{x'_n\}## such that ##x'_n=x_{n_0 + n-1}## is a subsequence of ##\{x_n\}## included in ##K(x_{n_0},1)## . By hypothesis, the ball is a complete subspace and the subsequence ##\{x'_n\}## is cauchy. But then ##\{x'_n\}## converges to some point ##x \in K(x_{n_0},1)## This implies ##\{x'_n\}## converges to the same point x in the metric space X. So the original sequence ##\{x_n\}## has a convergent subsequence that converges to the point x, this means that ##\{x_n\}## converges to x. I've proved that an arbitrary cauchy sequence in X is convergent, this implies every cauchy sequence in X is convergent, then X is a complete metric space.

Is my proof correct?