Is a topological space $(\mathbb{R}, \tau)$ normal, irrespective of the nature of $\tau$?
For convenience I post the definition of normality:
Definition A topological space $(X, \tau)$ is said to be normal, if for closed sets $C$, $D \subseteq X$, if $C \cap D = \emptyset$, then there exist open sets $U$ and $V$ such that $C \subseteq U$, $D \subseteq V$ and $U \cap V = \emptyset$
Using the definition above it seems to be the case for me that if disjoint subsets $C$ and $D$ of $\mathbb{R}$ can be found, a proof similar to this one can be given. I quote from the chosen answer of that post:
Every metric space is normal, in particular $\mathbb R^n$. The proof goes as follows:
For each $a\in A$, let $r_a=\frac{1}{3}d(a,B)$, and for each $b\in B$, let $s_b=\frac{1}{3}d(b,A)$. Now define $U=\bigcup_{a\in A}B(a,r_a)$ and $V=\bigcup_{b\in B}B(b,s_b)$. It is not hard to show that $A\subseteq U,$ $B\subseteq V,$ and $U\cap V=\varnothing$.
If no two such sets can be found, the statement would be vacuously true like in this case.
Am I missing anything crucial?