Yes, no matter which of the three definitions at your link is used. Suppose that $X$ is a $k$-space, and $Y$ is a closed subspace of $X$, and let $A\subseteq Y$; $A$ is closed in $Y$ iff $A$ is closed in $X$.
Using the first definition at your link: If $A$ is closed in $Y$, then $A$ is closed in $X$, so $A\cap K$ is closed in $X$ for each compact $K\subseteq X$, and hence $A\cap K$ is closed in $Y$ for each compact $K\subseteq Y$. If $A$ is not closed in $Y$, then $A$ is not closed in $X$, so $A\cap K$ is not closed in $X$ for some compact $K\subseteq X$. Let $K_0=K\cap Y$. Then $K_0$ is compact, $A\cap K_0=A\cap K$ is not closed in $X$, and $K_0\subseteq Y$.
Using the third definition at your link: If $A$ is closed in $Y$, then $A$ is closed in $X$, so $A\cap K$ is compact for each compact $K\subseteq X$, and hence $A\cap K$ is compact for each compact $K\subseteq Y$. If $A$ is not closed in $Y$, then $A$ is not closed in $X$, so $A\cap K$ is not compact for some compact $K\subseteq X$. Define $K_0=K\cap Y$. Then $K_0$ is compact, $A\cap K_0=A\cap K$ is not compact, and $K_0\subseteq Y$.
Using the second definition at your link: Suppose that there are a compact Hausdorff space $K$ and a continuous $f:K\to Y$ such that $f^{-1}[A]$ is not compact. Clearly $f$ can be viewed as a continuous map from $K$ to $X$, so $A$ is not closed in $X$, and hence $A$ is not closed in $Y$. If on the other hand $A$ is not closed in $Y$, then $A$ is not closed in $X$, so there are a compact Hausdorff space $K$ and a continuous $f:K\to X$ such that $f^{-1}[A]$ is not compact. Let $K_0=f^{-1}[Y]$. Then $K_0$ is closed in $K$ and hence compact and Hausdorff, and $f\upharpoonright K_0$ is a continuous map from $K_0$ to $Y$, such that $(f\upharpoonright K_0)^{-1}[A]=f^{-1}[A]$ is not closed in $K_0$.
