Consider the following two exercises in May's A Concise Course in Algebraic Topology.
(a) Any subspace of a weak Hausdorff space is weak Hausdorff. (b) Any closed subspace of a $k$-space is a $k$-space. (c) An open subset $U$ of a compactly generated space $X$ is compactly generated if each point has an open neighborhood in $X$ with closure contained in $U$.
A Tychonoff (or completely regular) space $X$ is a $T_1$-space (points are closed) such that for each point $x \in X$ and each closed subset $A$ such that $x \notin A$, there is a function $f: X \to I$ such that $f(x) = 0$ and $f(a) = 1$ if $a \in A$. Prove the following. (a) A space is Tychonoff if and only if it can be embedded in a cube (a product of copies of $I$). (b) There are Tychonoff spaces that are not $k$-spaces, but every cube is a compact Hausdorff space.
In view of these two problems, what should one mean by a "subspace" of a compactly generated space? We do not want to restrict the allowable subsets.
The most naïve idea is to define the subspace topology to be the k-ified relative topology, so it is no longer necessarily induced from the ambient space simply from intersections. But this doesn't really invoke the previous problems.
edit I think this is what May has in mind, since the k-ified subspace topology satisfies the relevant universal property; see the bottom of p.155 of this appendix.