Excuse me can you see this question Show that every seconed countable space is separable and first countable I tried on it but i am not sure , I get / Let(X,T) be seconed countable space so there exist B={ Bn : n belong to N } s.t B countable base for X . When we write B={ B1,B2, ...} It doesn't mean that B is countably infinite it means that either for some n belong to N B={B1,B2, ... , Bn} or B empty or B is countably infinite . If X non empty then B nonempty , if for some k belong to N Bk empty then {B1,B2, ... , Bk-1, Bk+1, ...} Is also basis for X so let us assume that Bn non empty for all n Let xn belong to Bn ( note that by axiom of choice there is a function f:N _ union of Bn s.t xn = f(n) which is belong to Bn and A = { x1,x2,...} That is A is subset of X , obtaining by Chossing one element from each mber of Bn . Here also it is quite possible that A is finite set . Claim : A is countable dense subset . Clearly A is countable since | A | less than or equal | B | . To prove A dense , take x belong to X and open set U containing x , now B is basis for X , U is an open set containing x but we have for any x belong to X there exist no belong to N s.t x belong to Bno , Bno subset of U and Bno intersection with A non empty . Therefore x belong to closure of A implies X subset of clouser of A thus A closure = X . Thus A is countable dense subset of X therefore X is separable space . To show that X is first countable . Let x belong to X , then choose the element of B that contains x , so Bx = { Bn : x belong to Bn } it is countable nhood base about x , thus X it is first countable .
Sorry for the rather long question. If is it rather confusing, let me know so I can clarify. I want to thank you in advance for taking the time to read this question.
If $X$ is second countable, it has a countable base $\mathscr B$ for its topology. Let $(B_i)_{i\in\Bbb N}$ be an enumeration of the elements of $\mathscr B$. For each $i$, choose $b_i\in B_i$. Then $\{b_i:i\in\Bbb N\}$ is dense. Thus $X$ is separable.
Secondly, since $\mathscr B$ is countable, each $x\in X$ has a countable neighborhood base. Thus $X$ is first countable.