Prove: if $X$ is a metric space, $A$ sequentially compact in $X$, then for any $a>0$ there are finite open balls of radius $a$ whose union covers $A$

Please prove: Suppose $X$ is a metric space, and $A$ is a sequentially compact set in $X$. If given a real number which is larger than zero (suppose $a>0$), there are finite open balls whose radius are $a$, and the union of these open balls can cover $A$. ($a$ is chosen randomly)

HINT: Suppose not; then for any finite $F\subseteq A$, the set $\{B(x,a):x\in F\}$ does not cover $A$. This means that we can recursively construct a sequence $\langle x_n:n\in\Bbb N\rangle$ in $A$ in the following way.
Let $x_0\in A$. Given $\{x_k:k<n\}$, we know that $\{B(x_k,a):k<n\}$ does not cover $A$, so we can choose $x_n\in A\setminus\bigcup_{k<n}B(x_k,a)$. In this way we construct a sequence $\langle x_n:n\in\Bbb N\rangle$ in $A$.
• Show that $d(x_m,x_n)\ge a$ whenever $m,n\in\Bbb N$ and $m\ne n$.
• Now use the sequential compactness of $A$ to get a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ converging to some $x\in A$ and derive a contradiction.
• This proof gives a little more: Namely that the centers of the open balls can all belong to $A$. – DanielWainfleet Jun 27 '16 at 2:46