My definition of $k$-space is:
Let be $(X,\tau)$ a topological subspace and $A\subset X$.
$A$ is a $k$-closed in $X$ if $\forall K\subset X$ compact it happens that $A\cap K$ is closed in $K$.
$X$ is a k-space if every $k$-closed of $X$ is closed in $X$.
My question is the following:
I was trying to prove that every closed subspace of a $k$-space is a $k$-space, but in my demonstration I do not use that subspace to be closed. If someone can gave me a clue that is wrong in my demonstration I would appreciate it.
Let be $(X,\tau)$ a $k$-space and $A\subset X$ closed, then $(A,\tau_{A})$ is a $k$-space.
Proof:
Let be $B\subset A$ a $k$-closed.
Then $\forall K\subset A$ compact it happens that $B\cap K$ is closed in $K$.
As $K\subset A \subset X$ and compactness is an absolute property, then $K$ is compact in $X$
$\Rightarrow$ $\forall K$ compact of $X$ it happens that $B\cap K$ is closed in $K$
$\Rightarrow$ $B$ is a $k$-closed of $X$ and $X$ is a $k$-space, then $B$ is closed in $X$.
Finally $\overline{B}^{A}=\overline{B}^{X}\cap A=B\cap A=B$
$\Rightarrow$ $B$ is closed in $A$ and therefore $A$ is a $k$-space
Obs: I mean absolute compactness for the following:
Theorem: Let be $(X,\tau)$ a topological space and $A\subset X$, then $B\subset A$ is compact in $A$ if and only if $B$ is compact in $X$.