question about compact and H closed ‎A subset S of a space ‎‎$‎(X, ‎\tau‎)‎‎$‎ is said to be quasi H-closed ‎subset ‎(resp. ‎subspace)‎ if for every cover ‎‎$ ‎\{ ‎V_i : i ‎\in ‎\alpha \}  ‎\subset ‎\tau‎$ ‎(resp.‎ ‎$ ‎\subset ‎\tau‎_{‎S }‎‎$)‎  ‎of ‎‎‎‎$‎S‎$‎ ‎‎‎, there exists a finite subset ‎$ ‎\alpha‎_{‎0‎}‎‎ ‎\subset‎ ‎‎\alpha‎‎‎‎ $‎ such that $ S ‎‎\subset ‎\bigcup‎_{‎i \in ‎\alpha‎_{0}‎‎‎}‎ cl‎_{‎\tau‎}‎(V_i)‎‎‎‎‎$ ( ‎resp. ‎$  S ‎‎\subset ‎\bigcup‎_{‎i \in ‎\alpha‎_{0}‎‎‎}‎ cl‎_{‎\tau‎_{‎S‎}‎‎}‎(V_i)‎$‎‎)‎.
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A space $(X, \tau)$ is said to be locally quasi H-closed space  if each point if X has a $\tau$ -open
neighbourhood which is a quasi H-closed subspace of $(X, \tau)$. Each quasi H-closed space is locally quasi H-closed, the converse is not true.
I want  to prove the following statement , but I am not sure is it right.  Can you give me help?
Quasi H-closed is compact iff every compact subset is  locally quasi H-closed.
 A: Let $(X,\tau)$ be a topological space (whether this is a subspace of a larger topological space doesn't matter for this). In general you have that any compact space is going to be quasi-H-closed. This is because given any open cover $\{U_i\}$ there's going to be a finite subcover and the closure of those elements still cover the space (or subspace). You also have that any quasi-H-closed space is going to be locally quasi-H-closed since you can take all of $X$ as the quasi-H-closed open neighborhood since $X$ is open and quasi-H-closed and contains all the points.
Summary: $\text{compact}\rightarrow \text{quasi-H-closed}\rightarrow \text{locally quasi-H-closed}$.
Now to address your question. For any kind of topological space it's compact sets will be locally quasi-H-closed. So the $(\rightarrow)$ implication is trivially true. For the $(\leftarrow)$ implication all one must find is a non compact space which is quasi-H-closed. Such an example is the real line with the topology generated by sets of the form $(r,\infty)$. This space is not compact since $( (-n,\infty))_n$ is an open covering that does not admit a finite subcover but it's quasi-H-closed since the closure of any set that doesn't have an upper bound is $\mathbb{R}$. This space is quasi-H-closed it trivially has the property that every compact subset is locally quasi-H-closed but the entire space is not compact.
Note: This is the first time I hear about quasi-H-closed spaces so I'm taking the definitions you provided at face value I hope this helps
Edit: the other question you asked is if the proposition was true for open sets instead of compact sets. This is still false, namely the $(\leftarrow)$ is not true. We reconsider $\mathbb{R}$ with the topology listed above. Every subset in that topology is locally quasi-H-closed since the closure of any non empty open set is the entire real line.
