Let $X$ locally compact and $\sigma$-compact. Then any open subset is also $\sigma$-compact Im stuck with this exercise

Show that if $X$ is locally compact and $\sigma$-compact then any open subset is also $\sigma$-compact.

Note: here locally compactness imply that the space is also Hausdorff.
This exercise is the last part of a longer exercise where the previous part was this.
If the linked exercise would be true (but it is not) then a locally compact and $\sigma$-compact space would be separable, and because I also know that open sets of locally compact spaces are also locally compact then the proof would be (I guess) easier to prove following this path.
However, as I said, the statement of $X$ being locally compact and $\sigma$-compact imply that it is also separable is not true in general. Then its possible that also this part of the exercise is wrong (that is, that the statement to be proved would be false).
In any case I dont find something to work to prove the statement. My first attempt was trying to show that if $X$ is Lindelöf, $\sigma$-compact and locally compact then any open set is also Lindelöf, but I dont found a way to show it, thus I'm lost again in the starting point.
The exercise appear (as stated in the linked question) in a book of analysis so it is supposed that it can be solved using elementary notions of topology.
Can someone confirm if the statement to be proved is really true or if, by the contrary, it is other mistake in the book? If it would be solvable, can someone give me a hint? Thank you.
 A: As in your previous question  you probably want to work in a metric space. Then this does hold, even without local compactness.
Let $X$ be  $\sigma$-compact metrisable and $O$ open in $X$.
In a metric space a closed set is a $G_\delta$ (we can write $F = \cap_n O_n$ where $O_n = \{x: d(x,F) < \frac{1}{n}\}$) and by taking complements and applying de Morgan we see that $O$ is an $F_\sigma$, so we can write $O = \cup_n F_n$ with $F_n$ closed in $X$.
As each $F_n$ is $\sigma$-compact, as a closed subset of a $\sigma$-compact space (holds in all spaces) and so $O$ is (a countable union of countably many compact subsets is still of that form).
Without an extra condition this does not hold: if $X$ is the lexicographically ordered square, $X$ is compact (so trivially $\sigma$-compact and locally compact) while its open subspace $[0,1] \times (0,1)$ is not $\sigma$-compact as it has an uncountable disjoint cover by open open sets $\{x\} \times (0,1), x \in [0,1]$.
A: The following three facts are known:
1) A locally compact space is $\sigma$-compact if and only if it is Lindelöf.
2) A Lindelöf space is hereditarily Lindelöf (i.e. every subspace is Lindelöf) if and only if all open subspaces are Lindelöf.
3) A Lindelöf space is hereditarily Lindelöf if and only if it is perfectly normal.
Now let $T = \Pi_{t \in I} I_t$ be the uncountable product of unit intervals $I_t$, $t \in I$. It is a compact space which is not perfectly normal. Hence it is not hereditarily Lindelöf so that there exists an open subspace $U$ which is not Lindelöf. Then $U$ is not $\sigma$-compact.
What can be said positively is that among locally compact $\sigma$-compact spaces the assertion holds precisely for perfectly normal $X$.
PS. Don't ask me for proofs of 1) - 3). These are exercises in the book "General Topology" by R. Engelking, and I have not done them.
A: Here's a simple counterexample. Let $X$ be an uncountable set, embedding in its one-point compactification $\alpha(X)$. Recall that the underlying set of this is $X \cup \{*\}$ where $* \not\in X$, and an open set is either a subset of $X$ or the complement of a finite subset of $X$ (including $*$).
Then the space $\alpha(X)$ is compact and Hausdorff, so is both locally compact and $\sigma$-compact, and has $X$, given the discrete topology, as an open subset. This is not $\sigma$-compact because compact subsets of discrete sets are finite, and $X$ was chosen to be uncountable.
This shows that not only does an open subset of a compact Hausdorff space not have to be $\sigma$-compact, there is no upper bound on the number of compact sets that we might need to cover it.
