Let $X=\mathbb Q^*=\mathbb Q\cup\{\infty\}$ be the one-point compactification of the rationals. The open sets in $X$ are the open sets in $\mathbb Q$, together with the complements in $X$ of the (closed) compact subsets of $\mathbb Q$.
I have not seen a reference for this, but it seems the following should hold:
The compactification $X=\mathbb Q^*=\mathbb Q\cup\{\infty\}$ is Frechet-Urysohn.
In other words, if $A\subseteq X$ is not closed in $X$ and $x\in\overline A\setminus A$, there is a sequence $(x_n)_{n\in\omega}$ in $A$ that converges to $x$. (see start of a proof below)
This is interesting because first countable $\implies$ Frechet-Urysohn $\implies$ sequential $\implies$ compactly generated. From the properties compiled here and the references there we already know $X$ is not first countable. Showing it's Frechet-Urysohn would imply it's compactly generated in the sense of Definition 2 in wikipedia. On the other hand, the product space $\mathbb Q^*\times\mathbb Q^*$ is not compactly generated. So $\mathbb Q^*$ would provide an example of a compactly generated space whose square is not compactly generated.
(Incomplete) Proof of result: Let's write $Y=\mathbb Q$ for convenience. Take $A\subseteq X$ with $x\in\overline A\setminus A$.
Case 1: Suppose $x\in \mathbb Q$. Then $\mathbb Q$ itself is an open nbhd of $x$ and $\mathbb Q$ is Frechet-Urysohn since it is first countable. So there is a sequence $(x_n)_n$ in $A\cap\mathbb Q$ that converges to $x$.
Case 2: $x=\infty$, so $A\subseteq \mathbb Q$. We have to show there is a sequence in $A$ that converges to $\infty$. By definition of the one-point-compactification, saying $\infty\in\overline A$ is equivalent to saying that $A$ is not contained in any compact subset of $\mathbb Q$. On the other hand, if we can find an infinite discrete subspace $Z\subseteq A$ that is closed in $\mathbb Q$, any enumeration of it would be a sequence converging to $\infty$. So the proof would boil down to this:
Suppose $A\subset\mathbb Q$ is not contained in any compact subset of $\mathbb Q$. Does $A$ necessarily contain an infinite discrete subspace $Z\subseteq A$ that is closed in $\mathbb Q$?
How to show this?
If $A\subseteq\mathbb{Q}$ is not contained in any compact subset of $\mathbb{Q}$ (i.e., its closure in $\mathbb{Q}$ is not compact), then there is a sequence $(a_n)$ in $A$ with no subsequence that converges in $\mathbb{Q}$. Such a sequence converges to $\infty$ in $\mathbb{Q}^*$ (any compact subset $K\subset\mathbb{Q}$ cannot contain a subsequence of $(a_n)$ so the sequence must eventually be outside of $K$).