Suppose a $T_4$ space $X$ satisfies a property that every continuous map $f$ from $X$ to any Hausdorff space $Y$ is a closed map. Show that $X$ is compact.
What I am thinking is to let $\mathcal{U}$ be an open cover of $X$ and find a way to choose finite open sets from $\mathcal{U}$ that can cover $X$ again. Do we need to define a suitable continuous map by ourselves and use the fact that it is a closed map to choose suitable open sets from $\mathcal{U}$? I have tried to define a continuous map $f \colon X \to Y$ as mapping different open sets from $\mathcal{U}$ to different points in $Y$. However, different open sets from $\mathcal{U}$ may intersect with each other and hence $f$ may not be well-defined. Apart form that, I have no idea about how to choose suitable $f$ and $Y$. Can someone help me please?
Edit 1: I am trying to apply the Urysohn Lemma or the Tietz Extension Theorem due to the normality of $X$. But still have no idea.
Edit 2: Finally found a way to prove without using compactification. I will post the proof when I’m free. Hint: Use Urysohn Lemma and Tychonoff Theorem.
Best Answer
Any such $X$ has a compactification $(\gamma X, e)$ in which $e: X \to \gamma X$ is an embedding with a dense image (this uses $T_{3\frac12}$). Then the assumption says that $e$ must be a closed map, so $\gamma X = e[X]$ and $X$ is compact.