Quoting the wikipedia entry for regular space
> A topological space $X$ is a _regular space_ if, given any closed set $F$ and any point $x$ that does not belong to $F$, there exists a neighbourhood $U$ of $x$ and a neighbourhood $V$ of $F$ that are disjoint.
If the topology on $X$ is indiscrete, there are only two cases to consider: $F = \emptyset$ and $F = X$. In the first case, you can take $U = X$ and $V = F = \emptyset$. Then $x \in U$ and $U \cap V = \emptyset$. In the latter case, there is no $x$ that does not belong to $F$ and the result is trivial.
The proof for normality is similar.