The statement is accurate, but must be understood correctly: in every group $G$ there is a subgroup $D \leq G$ with the following properties: (1) $D$ is perfect, (2) if $H \leq G$ and $H$ is perfect, then $H \leq D$, and (3) $D$ is normal.
By (1) and (2) there is only one such subgroup, so (3) is not surprising.
If $G$ is cyclic of prime order (or abelian, or solvable), then $D=1$ is the trivial subgroup. This is the only perfect subgroup of $G$, so it satisfies (2) in an unimpressive way.
If $G$ is non-abelian simple, then $G$ itself is perfect, so $G \leq D$ by (2), but since $D \leq G$, we get $D=G$.
If $G$ is finite, then one can find $D$ using Jared's Hint #1 and Maths Lover's observation in the comments: If $H \leq G$ then $[H,H] \leq [G,G]$. Notice that $D=[D,D] \leq [G,G]$ and when we try to find the $D$ for $G$, we would get the same as if we tried to find the $D$ for $[G,G]$. Hence we just keep replacing $G$ with $[G,G]$, until finally $G$ and $[G,G]$ are equal. At that point they are equal to $D$. This stops after a finite number of steps in a finite group. [ For an infinite group the same idea works, but at limit ordinals you take intersections, and the overall process is transfinite induction. ] This is called the “residual” method, and writes $D$ as the solvable [ or hypoabelian ] residual.
However, Dummit–Foote outline a different method, the radical method. If $A,B$ are perfect subgroups then set $C=\langle A,B \rangle$. Then $C = \langle A , B \rangle = \langle [A,A], [B,B] \rangle \leq [C,C] \leq C$, so $C=[C,C]$ is also perfect. $D = \langle A \leq G : A = [A,A] \rangle$ is thus a perfect subgroup of $G$.