= Lexicographic order topology on the real plane
Give $\mathbb R^2$ the <lexicographic order> and the <order topology>. Every point $(a,b)$ has an interval neighbourhood between $(a,b-\varepsilon)$ and $(a,b+\varepsilon)$, lying entirely in the vertical line $\{a\}\times\mathbb R$. These intervals give each such line its usual real <topology>, and every line is open. Thus this space is a disjoint union of uncountably many open copies of the real line and is not a <second-countable space>: any <basis of a topology> must supply a different member inside each of these disjoint nonempty open sets.
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