Is it true that, in a topological space $(X, \mathcal{T})$, regularity does not imply normality and vice versa?
I looked for examples to prove this; but I just don't know many examples to look into. If it is really true, can anyone give one or two good examples for both the cases? Regards.
I presume you're working with the definitions that say that a space is regular/normal if a point and a closed set/two closed sets can be separated by open neighbourhoods.
An example of normality not implying regularity is the excluded point topology. For concreteness, set $X=\mathbb{N}$ and let the open sets be $\mathbb{N}$ and any subset not containing 0. Then all nonempty closed sets contain 0, so the space is normal vacuously, but it cannot be regular, since any open neighbourhood of a closed set must in fact be the whole space.
Note, however, that normality implies regularity if your space is $T_1$.
A regular space which isn't normal is slightly trickier to cook up. The following is a slightly simplified presentation of what is usually called the deleted Tychonoff plank. Let $A$ be a discrete countable space, $B$ a discrete uncountable space and $A^*,B^*$ their one-point compactifications. Let $X=(A^*\times B^*)\setminus\{(\infty_A,\infty_B)\}$. Since both $A$ and $B$ are locally compact and Hausdorff, $A^*$ and $B^*$ are Hausdorff, so $A^*\times B^*$ is compact and Hausdorff. Since $X$ is an open subspace of the product, it is locally compact and Hausdorff. Hence, by a standard result, it is regular.
To see that $X$ isn't normal, you can try to prove that the sets $M=\{(\infty_A,b);b\in B\}$ and $N=\{(a,\infty_B);a\in A\}$ are disjoint closed sets with no disjoint open neighbourhoods.