Sigma-compactness if there exists an open cover ... $X$ is locally compact Hausdorff space.
Show that $X$ is $\sigma$-compact if there is an open cover $\{A_n\}$ of $X$  (i.e $X=\bigcup_{n=1}^\infty A_n$)
such that $\operatorname{Cl}(A_n)$ is compact and $\operatorname{Cl}(A_n) \subset A_{n+1}$ for all $n\\$.
$\\$
Here is the sketch of my proof.
Since $X$ is locally compact Hausdorff, there exists an open set $V_n$ whose closure is compact such that $\operatorname{Cl}(A_n) ⊂ V_n ⊂ \operatorname{Cl}(V_n) ⊂ A_{n+1}$
Let $Q_n = \bigcup_{j=1}^n \operatorname{Cl}(V_j).$
Then $Q_n$ is compact and $X=\bigcup_{j=1}^\infty Q_n$
Thus, $X$ is sigma compact.
Any help would be appreciated.
Thank you.
 A: It's in fact an equivalence. 
The easier direction is even easier than you showed: just note that $X = \bigcup_n \operatorname{Cl}(A_n)$ and thus a countable union of compact sets and thus $\sigma$-compact. The $Q_n$ are not needed, as being increasing is not a requirement of $\sigma$-compactness (and moreover the $A_n$ and thus $Q_n$ already were increasing so the union is superfluous). 
For the other direction, write $X$ as an increasing union (using the finite unions idea, indeed) of compact sets $K_n$ and apply the lemma that in a locally compact Hausdorff space we have that for each compact $K$ and open set $U$ with $K \subseteq U$ we have an open set $V$ such that $\operatorname{Cl}(V)$ is compact with $K \subseteq V \subseteq \operatorname{Cl}(V) \subseteq U$ for each of those $K_n$ and use the $V_n$ from the lemma.
Note that the existence of the $A_n$ already implies the local compactness too. 
So we have for a Hausdorff space $X$ that the following are equivalent:


*

*$X$ is locally compact and $\sigma$-compact.

*There exists such an increasing cover of open sets $A_m$ with compact closure as in the question.

