Second Countable, First Countable, and Separable Spaces

Upon further studying Topological Spaces, I understand:

• If a space $X$ has a countable dense subset, then $X$ is a separable space.
• A space $X$ is first countable provided that there is a countable local basis at each point of $X$.
• A space $X$ is second countable if and only if its topology of $X$ has a countable basis.

My question is:

Why would a space $X$ that is second countable also be first countable and separable? Why would a space $X$ that is first countable not necessarily be considered a separable space and vice versa?

I do have a rough idea as to why a second countable space is also a first countable space.

A second countable space has a countable basis $\mathcal{B}$ $-$ which consist of a countable family of open sets $-$ then the members of $\mathcal{B}$ which contain a particular point $a$ form a countable local basis at $a$. Thus each second countable space is first countable.

Now if the space $X$ is second-countable, to also be separable, there needs to exists a countable dense subset of $X$. It has been established that if $X$ is a second countable space, it has a countable basis $B$. This is where I get stuck.

I am not sure as to why spaces that are first countable do not imply they are separable and vice verse. Does it have to do with the fact first countable spaces deal with countable local basis that may or may not be dense?

Am I on the right track?

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. I greatly appreciate any assistance you provide.

Your proof that a second countable space $X$ is first countable is correct. To show that it’s also separable, let $\mathscr{B}$ be a countable base for it. For each $B\in\mathscr{B}$ let $x_B\in B$; clearly $\{x_B:B\in\mathscr{B}\}$ is a countable subset of $X$, and it’s pretty straightforward to show that it’s dense in $X$.

For the negative results, let $\tau$ be the discrete topology on $\Bbb R$. Then $\langle\Bbb R\tau\rangle$ is first countable, because for each $x\in\Bbb R$ the finite set $\big\{\{x\}\big\}$ is a local base at $x$. Now show that the only dense subset of $\Bbb R$ in this topology is $\Bbb R$ itself; since $\Bbb R$ is uncountable, the space cannot be separable, and therefore, of course, it can’t be second countable, either.