A permutation group $G$ acting on $X$ (that is, a subgroup of $S_X$) is regular if and only if for ever $x\in X$, the stabilizer of $x$ is trivial; that is, the stabilizer of $x$, $G_x = \{g\in G\mid gx = x\},$ equals the trivial group for every $x\in X$.
If $g\in G$ is such that $gx = y$, then it is an easy matter to verify that $G_x = gG_yg^{-1}$; in particular, if the action of $G$ is transitive, then $G$ is regular if and only if $G_x=\{1\}$ for at least one (and hence for all) $x\in X$.
Given a group $G$ and a subgroup $H$ of $G$, the core of $H$ is the largest normal subgroup of $G$ that is contained in $H$; as you note, this is equal to the intersection of all conjugates of $H$ in $G$. A subgroup $H$ of $G$ is core-free if the core of $H$ in $G$ is the trivial group.
Your inference that showing a subgroup is core-free is the same as showing that core is the trivial subgroup is correct, modulo a small caveat; corrected: However, the professor is saying you should show that there are no core-free proper subgroups, so you need to show that the core of any subgroup (other than the trivial group) is not the trivial group.