Is this property of continuous maps equivalent to properness? For the purposes of my question, a continuous map $f : X \to Y$ is proper if it is closed and the preimage of every compact subspace of $Y$ is a compact subspace of $X$.
Say a continuous map $f : X \to Y$ is semiproper if, for every continuous map $y : T \to Y$ where $T$ is compact, the space $T \times_Y X = \{ (t, x) \in T \times X : y (t) = f (x) \}$ is compact.
It is a fact that a closed map is proper if and only if it is semiproper.
Question. Are semiproper maps always closed?

If $Y$ is a compactly generated Hausdorff space, then it is easy to check that every semiproper map $f : X \to Y$ is closed – indeed, we only need the defining property for subspace inclusions $y : T \to Y$. On the other hand, if we weaken the definition by restricting to subspace inclusions $y : T \to Y$, then there are easy counterexamples. 
That leaves non-(compactly generated Hausdorff) spaces. Perhaps there is a counterexample there?
 A: Let $X$ be an uncountable discrete space, let $Y$ be the same set with the cocountable topology, and let $f:X\to Y$ be the identity map.  Then $f$ is not closed, but I claim $f$ is semiproper.  Indeed, if $T$ is compact and $y:T\to Y$ is continuous, then the image of $y$ must be compact in $Y$ and hence finite.  The topologies of $X$ and $Y$ agree on finite sets and so $T\times_Y X\cong T\times_Y Y\cong T$ is compact.  (Explicitly, $T\times_Y X$ is just $T$ with its topology refined so that each fiber of $y$ is clopen, but each fiber of $y$ is already closed in $T$ and thus clopen since there are only finitely many of them.)
More generally, let $Y$ be any space and let $X$ be its CG-ification (i.e., $X$ is $Y$ with the topology generated by its compact subspaces).  Then every continuous map $y:T\to Y$ from a compact space is also continuous as a map to $X$, and so $T\times_Y X\cong T$ is compact (since $T\times_Y X$ is just $T$ with its topology refined so that $y$ is continuous as a map to $X$).  So the identity map $X\to Y$ is semiproper, but is not closed unless $Y$ is compactly generated.
