Prove that $X$ is a complete metric space if every closed ball in $X$ is compact.

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Let $(X, d)$ be a metric space.

Let $x_n \in X$ be a Cauchy sequence.

Proposition 1 $$(x_n) \text{ Cauchy} \Longrightarrow (x_n) \text{ Bounded}$$

Proposition 2 $$(x_n) \text{ Bounded} \Longrightarrow \exists M>0: \forall n,o\in \mathbb{N},\; d(x_n,x_o)\leq M$$

Therefore $x_n\in\bar{B}(x_o,M). $

Proposition 3 $$\bar{B}(x_o,M) \text{ Compact} \Longrightarrow \exists x_{n_k} \to a\in \bar{B}(x_o,M)\subset X$$ Where $x_{n_k}$ is a subsequence of $x_n$ with $k\in\mathbb{N}$.

Proposition 4 $$(x_n) \text{ Cauchy} \wedge (x_{n_k}) \text{ Convergent} \Longrightarrow (x_n) \text{ Convergent}$$

Therefore $(X,d)$ is complete.

Somehow I feel this is too good to be true, and I do not feel confident about my proof. Any comments, corrections, or insight would be greatly appreciated.