A metric space is separable iff it is second countable [closed]

HINT: One direction is pretty trivial; I’ll leave it to you, at least for now. The harder direction is to prove that a separable metric space is second countable. Suppose that $\langle X,d\rangle$ is a separable metric space. Then $X$ has a countable dense subset $D$. The set $\Bbb Q$ of rational numbers is countable, so

$$\mathscr{B}=\{B(x,r):x\in D\text{ and }0<r\in\Bbb Q\}$$

is a countable family of open balls in $X$. (Here $B(x,r)=\{y\in X:d(x,y)<r\}$ is the open ball of radius $r$ centred at $x$.) Show that $\mathscr{B}$ is a base for the metric topology on $X$. In other words, show that if $U$ is a non-empty open set in $X$, and $x\in U$, then $x\in B\subseteq U$ for some $B\in\mathscr{B}$. You will find it helpful to note that there is some $\epsilon>0$ such that $B(x,\epsilon)\subseteq U$; thus, you need only show that there is some $B(y,r)\in\mathscr{B}$ such that $x\in B(y,r)\subseteq B(x,\epsilon)$. The triangle inequality will be helpful.

For one implication you have a countable base of your space. Take a point in every element of the base. Since every open set is union of elements of the base, every open sets contains one of the choosen points...

Conversely, if you have a countable dense set, consider the family of all open balls with rational radius centered in your set. This family is countable, and it is easy to show that it is a base.