Does every compact metric space have a finite basis?

Question

Let $M$ be a compact metric space. Then the collection $\mathscr{U}=\{B(x,\frac1n):x\in M, n\in\Bbb{N}\}$ forms an open cover of $M.$ Let $\mathscr{B}_0$ be a finite sub-cover extract from $\mathscr{U}$. We can easily prove that $\mathscr{B}_0$ is a (finite) sub-basis for the topology on $M$.
Does $\mathscr{B}_0$ provide a finite basis for the metric topology on $M$?

Answer 1

From the comments it appears that what you really want to show is that a compact metric space $X$ is second countable. To do that, for each $n\in\Bbb Z^+$ let $\mathscr{U}_n=\left\{B\left(x,\frac{1}n\right):x\in X\right\}$; $X$ is compact, so $\mathscr{U}_n$ has a finite subcover $\mathscr{B}_n$. Let $\mathscr{B}=\bigcup_{n\in\Bbb Z^+}\mathscr{B}_n$; then show that $\mathscr{B}$ is a base for $X$. This takes a little work with the triangle inequality but isn't too hard.