Weak Hausdorff space

Question

A topological space $X$ is weak Hausdorff if for any compact Hausdorff space $K$ and every continuous map $f:K \to X$, the image $f(K)$ is closed in $X$. Is $f(K)$ a Hausdorff subspace of X??

Answer 1

Yes, essentially because $f:K \to f[K]$ is a perfect map and then Hausdorffness is preserved by forward images. We also use that $X$ is $T_1$ for that; see this answer for more details.