If $f$ is a function from $\Bbb R$ in the indiscrete topology to $\Bbb R$ in the usual topology then $ f$ is continuous iff it is constant. A constant map is continuous between all spaces, and for the reverse we just need to note that if $f$ has at least two distinct values, say $p$ and $q$, then $f^{-1}[\\{p\\}]$ and $f^{-1}[\\{q\\}]$ are disjoint closed non-empty subsets (by continuity and standard properties of inverse images) and in the indiscrete topology that cannot happen: there is only one non-empty closed set, namely the whole space. So a non-constant map is not continuous.
Any map in the other direction ( so to the indiscrete topology) is always continuous.