If $X^{\star} \times Y^{\star}$ is $KC$ space, then $X\times Y$ is a $k$-space

Question

The following theorem is found in the article "ON KC AND k-SPACES, A. García Maynez. 15 No. 1 (1975) 33-50"

Theorem 3.5: Let $X, Y$ be topological spaces. If $X^{\star} \times Y^{\star}$ is KC then $X\times Y$ is a $k$-space

Definitions:

1. $X$ is a $KC$ space if every compact $K\subset X$ is closed.
2. Let $A⊂X$. Then $A$ is a $k$-closed if for all compact $K⊂X$ it happens that $A∩K$ is closed in $K$
3. $X$ is a $k$-space if every $k$-closed set of $X$ is a closed set in $X$.
4. Let $C\subset X$. Then $C$ is compactly closed if for all compact and closed $K\subset X$, it happens $C\cap K$ is compact.

The article refers that the theorem 3.5 is a consequence of theorem 3.4

Theorem 3.4: Let $X,Y$ be non-compact spaces and let $Y^{\star}=Y\cup \{ \infty \}$ be the one-point compactification of $Y$. Assume $Y$ is a $KC$ space. Then a set $C\subset X\times Y$ is compactly closed in $X\times Y$ if and only if $C\cup (X\times \{ \infty \})$ is compactly closed in $X\times Y^{\star}$.

My attempt: Let $A\subset X\times Y$ $k$-closed. I want to show that $A$ is closed in $X\times Y$.

By hypothesis and $KC$ is a hereditary property, we have to $X\times Y$ is $KC$ space. Also $KC$ is a factorizable property, then $Y$ is a $KC$ space. Now, clearly every $k$-closed set is compactly closed, therefore $A$ is compactly closed in $X\times Y$. Then by the theorem 3.4 we have $A\cup (X\times \{\infty \})$ is compactly closed in $X\times Y^{\star}$.

Let $K\subset X\times Y$ compact and closed. Consider the projection function $\rho_2 : X\times Y \rightarrow Y$, which is continuous and by the compactness of $K$, then $\rho_2 (K)$ is compact and closed in Y, beacuse $Y$ is $KC$. Note that $Y-\rho_2 (K)$ is open in $Y^{\star}$ and $Y^{\star}-\rho_2 (K)=Y^{\star}\cap (Y-\rho_2 (K))$, which is open in $Y^{\star}$, therefore $\rho_2 (K)$ is closed in $Y^{\star}$ and $X\times \rho_2 (K)$ is closed in $X\times Y^{\star}$. As $K\subset X\times \rho_2 (K)$ we have $K$ is closed in $X\times Y^{\star}$

Now remember $A\cup (X\times \{\infty \})$ is compactly closed in $X\times Y^{\star}$ it happens that $A\cap K=K\cap (A\cup (X\times \{\infty \}))$ is compact in $X\times Y$. Therefore $A\cap K$ is closed in $X\times Y$ because $X\times Y$ is KC.

I still don't have a clear idea of how to conclude that $A$ is closed in $X\times Y$. I hope you can help me or make any observations of my proof

Thank you.