Prove or disprove: Every normal space is a regular space.
Disprove by the following example
$X=\{a,b,c\}$ and $\tau=\{∅,X,\{a\} \}$
The problem is how to show this example is a normal space ?
Definition of Normal space: a space $X$ is called normal space iff for every pair $L$ , $M$ of disjoint closed sets in $X$ there exists open sets $G$ and $H$ s.t $L⊂G$ and $M⊂H$ and $G ∩ H = ∅$
The closed sets of $X$ are the complements of the open sets, hence we have three of them: $\emptyset, X, \{b,c\}$. The only pairs of disjoint closed subsets occur when we have one of them be $\emptyset$ (as the other $2$ intersect), and in that case we can choose $\emptyset$ as its open neighbourhood, and $X$ for the other one (and these are disjoint open neighbourhoods). This shows that $(X,\tau)$ is normal.
On the other hand, $a \notin \{b,c\}$ and the latter set is closed as we saw. The only open neighbourhood of $\{b,c\}$ is $X$ and this intersects any neighbourhood of $a$ (in at least $a$ itself). So we cannot separate the point $a$ and the closed set $\{b,c\}$ by disjoint open neighbourhoods. This shows that $(X,\tau)$ is not regular.