A criteria for an open subset of a compactly generated space to be compactly generated Is the following statement true or false?

An open subset $U$ of a compactly generated space $X$ is compactly generated if each point of $U$ has an open neighborhood in $X$ with closure contained in $U$.


(Added 2/16/2023) Summary of the answers:
The answer depends on which notion of "compactly generated" (CG) space is used.  There are multiple variants of the concept, as mentioned here for example, or in Stefan Hamcke's comment.
Definition 1: $X$ has the final topology with respect to the inclusions from all its compact subspaces.  This is the definition in wikipedia.
Definition 2: $X$ has the final topology with respect to all continuous maps from arbitrary compact Hausdorff spaces.  This is the definition in nlab and is the one more commonly used in algebraic topology.
Eric Wofsey's answer: with Definition 1, the desired statement is true.  Also note as a consequence:

(With definition 1)  In a CG regular space, all open subspaces are CG.

David Carchedi's answer: with Definition 2, all the open subsets of a CG space are CG.
 A: ANY open subset of a compactly generated (possibly non-Hausdorff) space is compactly generated. Indeed, let $U$ be an open subset of a compactly generated space $X.$ Let $A$ be a subset of $U$ such that for any continuous map $f:K \to U$ from $K$ compact Hausdorff, $f^{-1}\left(A\right)$ is open in $K$. We want to show that $A$ is open in $U.$ But as $U$ is open in $X,$ this is iff $A$ is open in $X.$ So let $g:C \to X$ be a continuous map from a compact Hausdorff space. It suffices to show that $g^{-1}\left(A\right)$ is open in $C.$ Now, $g^{-1}\left(U\right)$ is an open neighborhood of $g^{-1}\left(A\right)$ in $C$ so it suffices to show that $g^{-1}\left(A\right)$ is open in $g^{-1}\left(U\right)$. But, as an open subspace of a compact Hausdorff space, $g^{-1}\left(U\right)$ is locally compact Hausdorff, and hence compactly generated. So it suffices to prove that for any map $h:D \to g^{-1}\left(U\right)$ with $D$ compact Hausdorff, $h^{-1}\left(g^{-1}\left(A\right)\right)$ is open in $D.$ But $$D \stackrel{h}{\to} g^{-1}\left(U\right) \stackrel{\tilde g}{\to} U$$ is a continuous map to $U$ from a compact Hausdorff space, where $\tilde g$ is the restriction of $g,$ so by assumption, $h^{-1}\left(\tilde g^{-1}\left(A\right)\right)$ is open in $D.$ But as $g^{-1}\left(U\right)$ contains $g^{-1}\left(A\right),$ this is the same as $h^{-1}\left(g^{-1}\left(A\right)\right),$ finishing the proof.
A: This is true.  Suppose $A\subseteq U$ is closed with respect to the compactly generated topology on $U$ and let $V=U\setminus A$.  We wish to show that $V$ is open in the subspace topology on $U$, or equivalently that $V$ is open in $X$.  Let $K\subseteq X$ be compact; it suffices to show that $V\cap K$ is open in $K$.  Fix a point $x\in V\cap K$ and let $W\subseteq X$ be an open neighborhood of $x$ whose closure is contained in $U$.  Then $L=\overline{W}\cap K$ is closed in $K$ and hence compact, and is contained in $U$.  Since $A$ is closed with respect to the compactly generated topology on $U$, $A\cap L$ is closed in $L$, and so $V\cap L$ is open in $L$.  Since $x\in V\cap L$ and $L$ contains a neighborhood of $x$ in $K$ (namely $W\cap K$), $V\cap L$ contains a neighborhood of $x$ in $K$, and hence so does $V\cap K$.  Since $x\in V\cap K$ was arbitrary, this means $V\cap K$ is open in $K$.
