Let $(X,d)$ be a metric space, and let $K\subseteq X$. Suppose that every cover by open balls of $K$ has a finite sub-cover. I need to prove that $K$ is sequentially compact.
I thought on trying to take an arbitrary open cover, and maybe somehow use the fact that every open subset can be represented by a union of open balls, but that failed. Any hints?
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Show first that any infinite countable set $A$ in $X$ has a limit point. If not, every $X \in X$ is not a limit point which means that every $x \in X$ has an open neighbourhood $O_x$ such that $O_x \cap A \subseteq \{x\}$. THis gives an open cover of $X$, which has a finite subcover $O_{x_1},\ldots O_{X_N}$. But then
$$A = A \cap X \subset A \cap (\cup_{i=1}^N O_{x_i}) = \cup_{i=1}^N (O_{x_i} \cap A) \subseteq \{x_1, \ldots x_N\} $$
which implies $A$ is finite, a contradiction.
(this is the general proof that a compact set is limit point compact).
Next, if $(x_n)$ is a sequence in $X$, suppose that $A = \{x_n: n \in \mathbb{N}\}$ is infinite, otherwise some value $x_k$ occurs infnintely many often, yielding a constant (hence convergent) subsequence and we'd be done. But then there is some limit point $p $ of $A$, i.e. a point $p$ such that for every $\varepsilon >0$, $B(p,\varepsilon) \cap A$ has some point unequal to $p$. In fact every ball intersects $A$ in an infinite set.
This allows one to construct a subsequence converging to $p$: Pick $n_1$ with $x_{n_1} \in B(p,1)$. Pick $n_1 > n_2$ with $x_{n_2} \in B(p, \frac{1}{2})$ and $n_2 > n_1$ with $x_{n_3}\in B(p, \frac{1}{3}$, etc.
THis argument works for any first countable $T_1$ space.
You want to show that every sequence in $K$ has a limit point. Suppose the sequence $(a_n)$ has none. For each $x\in K$, there exists $\delta_x$ such that the sequence $(a_n)$ visits $B(x,\delta_x)$ only finitely many times.
The set of balls $\{B(x,\delta_x):x\in K\}$ is an open cover of $K$ by balls.
Can you finish and fill in the details?