Characterization of compact metrics spaces

Question

Let $X$ be a metric space. If every infinite subset of $X$ has at least one accumulation point, then $X$ is compact.

Does anyone know how to prove that?

Answer 1

Take a sequence $(a_n)_{n\in\Bbb N}$ in $X$. Then $A:=\{a_n~\vert~n\in\Bbb N\}$ is an infinite subspace of $X$ so it must have an accumulation point $x\in X$. For $n\geq 1$, $A$ must intersect $B(x,1/n)-\{x\}$. Take $\varphi(n)\in\Bbb N$ such that $$a_{\varphi(n)}\in B(x,1/n)-\{x\}$$ and do this by induction to make sure that $\varphi$ is strictly increasing. Then $(a_{\varphi(n)})_{n\in\Bbb N}$ is a subsequence of $(a_n)_{n\in\Bbb N}$ converging to some limit $x$, so $X$ is compact.

Answer 2

$X$ is totally bounded. That follows from the argument in this question and also, $X$ is complete, because a Cauchy sequence (its set of values, rather) will have an accumulation point, hence a convergent subsequence and so it is itself convergent to that point. And a totally bounded complete metric space is compact.