How to show that a regular space with non-countable basis is not normal I read that every regular space $X$ with a countable basis is normal. For if $A$ and $B$ are disjoint closed sets in $X$, one can form a neighborhood $U=\bigcup U_j$ over $A$ (from countable bases $A_j$) and another neighborhood $V=\bigcup V_j$ over $B$ (from countable bases $V_j$) s.t. $U$ and $V$ are disjoint, which makes $X$ normal. But since this proof relies on countable bases $U_j$ and $V_j$ to show normality, how can we show that if the second countability axiom is not met (i.e. the bases are uncountable), then $X$ is not normal.
 A: There are plenty of regular and normal spaces that are not second countable.
(e.g. an uncountable discrete space, the Sorgenfrey line/lower limit topology, the lexicographically ordered square, $[0,1]^I$ for $I$ uncountable etc.)
The theorem you read is indeed true: if $X$ is regular, we cannot normally conclude it is normal (there are quite a few examples for that too), but we can if we know additionally that $X$ is second countable (and then we can even go on to say it's metrisable by Urysohn's theorem, at least if regular implies $T_0$ or $T_1$)
But there are many ways a space can be both regular and normal without being second countable. E.g. being a metric or an ordered space. So if $X$ is not second countable, it doesn't necessarily mean anything on its status as a regular or normal space. 
The quoted theorem is often stated as a lemma in the proof of Urysohn's theorem (in order to show metrisability). It's not an important fact in its own right I'd say. (Having normality then gives Urysohn functions etc.)
