almost closed mapping Definition: The mapping$f:X\to Y$is almost closed if for every regular-closed set of $X$ like $F$, $f(F)$ is closed in $Y$.
Definition: A ‎subset A ‎of ‎‎space ‎(X,τ) ‎has $\alpha$-‎property if ‎every $\tau$‎‎-open cover of ‎‎‎A‎‎ has a finite subfamily whose $\tau$‎‎‎-closures cover ‎‎A‎.

Let $f:X\to Y$  be a  continuous mapping and $Y$ is Hausdorff space.then  $X$ ‎has $\alpha$-‎property if only if the mapping $f$ is almost closed.

My proof for one part:
suppose  $X$ ‎has $\alpha$-‎property and $F\subset X$ is regular-closed. I know that every  regular-closed subset of a space $X$ that has $\alpha$-‎property,has also $\alpha$-‎property. The $\alpha$-‎property  is reserved  under continuous mapping, so  $F(f)$ has  $\alpha$-‎property and in Hausdorff space is closed. Then f is almost closed.
I have a problem with the other side and I can not prove that $X$ has  $\alpha$-‎property. Can you help me?
 A: The reason why you couldn't get the other direction is because it is not true.
From your proof of the direct implication, I assume you already know that when a space $X$ satisfies the $\alpha$-property, then $X$ is called an $H$-closed space. An excellent book that deals with this kind of spaces is Porter & Woods - Extensions and Absolutes of Hausdorff Spaces; particularly, check section 4.8.
For a counterexample to the direction that you couldn't get, consider the identity map $\text{Id}:\mathbb{R}\to \mathbb{R}$, where $\mathbb{R}$ is endowed with the usual Euclidean metric. It is clear that $\text{Id}$ is an almost closed continuous map with a Hausdorff codomain. However, if we consider the open cover $\mathcal{U}=\{(-n,n):n\in\mathbb{N}\}$, and let $\mathcal{V}$ be a finite and non-empty subfamily of $\mathcal{U}$, then there must exist $m\in \mathbb{N}$ in such a way that $(-m,m)=\bigcup\mathcal{V}$. So, if we take closures in both sides of the equality, we get $[-m,m]=\overline{ \bigcup\mathcal{V}}$. It follows that $\overline{ \bigcup\mathcal{V}}$ cannot cover $\mathbb{R}$.
