Does a separable metric space have a countable basis?

Question

I want to prove that if $X$ is a metric space and has a dense countable subset, then it has a countable basis. I know that every metric space is first countable, but I can't continue. Thanks for your help.

Answer 1

Try balls of radius $1/m$ centred at the points $x_n$ of your dense countable subset.

Edit: to see how this works, consider a ball of radius $r$ around any point $p$. What conditions on $m$ and $x_n$ would guarantee that the ball of radius $1/m$ around $x_n$ contains $p$ and is contained in this ball?

Answer 2

Hint: Well, you have a countable collection of points to use as centers of balls, and you have a countable basis $\{B(x,\tfrac1n):n\in\mathbb N\}$ at each such point $x$, so...

Answer 3

Well, take a dense subset of $D\subset X$ and then the metric guarantees you a countable neighbourhood basis at every point of $D$. A countable union of countable sets is again countable. Check that this is enough.