Are countable spaces (i.e. $\mathbb{N}$ with any topology) second-countable? A countable space can have at most $2^\omega$ open subsets which suggests that a counterexample may exist. On the other hand both discrete and anti-discrete (or more generally with countable topology) spaces are second-countable. Also note that obviously a countable space is separable. So if it is additionally metrizable then it is second-countable.
But I couldn't prove that in general. Or is there a counterexample?
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$\begingroup$Consider $ω$ many convergent sequences, and glue their limits. The resulting space won't have countable base at the common limit point.
Also note that a countable space is second-countable if and only if it is first-countable.
$\endgroup$ 5 $\begingroup$Take a free ultrafilter on $\mathbb{N}$ and add $\emptyset$ to it to make it a topology. Standard facts on ultrafilters tell us that this is not second countable.
Another advanced example :let $x \in \omega^\ast$ and let $X= \omega \cup \{x\}$ in the subspace topology.
Or let $D$ be a countable dense subset of $\{0,1\}^{\Bbb R}$ in the product topology. I wrote extensively on such spaces here.
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