I want some confirmation about the correctness of the proof below. It seems correct but Im not totally sure if I had overlook something.
Let $(X,\mathcal T)$ a Hausdorff topological space, and $(K_j)\in X^{\Bbb N}$ a sequence of compact sets such that $X=\bigcup_{j\in\Bbb N} K_j$. Then for $$\mathcal K:=\{K\subset X: K\text{ is compact}\}$$ show that ${\cal A_\sigma(K)=B}(X)$.
Note: here $\mathcal B(X)$ is the Borel $\sigma$-algebra and $\mathcal A_\sigma(S)$ is the $\sigma$-algebra induced by some $S\subset\wp(X)$.
Observe that if $K\in\cal K$ then, cause $X$ is Hausdorff, $K$ is closed so $K^\complement \in\cal T$ and setting $\mathcal K^\complement:=\{K^\complement\subset X:K\in\mathcal K\}$ is clear (by definition of induced $\sigma$-algebra) that $\cal A_\sigma(K^\complement)=A_\sigma(K)$. And because $\cal K^\complement \subset T$ then we find that $\mathcal{A_\sigma(K)}\subset \mathcal B(X)$.
To show the other inclusion first note that the sequence $(K_j)$ is a countable compact cover of $X$, and any compact set in a Hausdorff space is closed, then every closed set $C$ have $(K_j)$ as a countable compact cover and $C\cap K_j$ is compact because is closed (by definition of topology) and $C\cap K_j\subset K_j$ (that is, its known that any closed subset of a compact space is compact).
Then the sequence $(C_j)$ where $C_j:=C\cap K_j$ is a (countable) sequence of compact sets with the property that $\bigcup C_j=C$. Because this holds for any closed set of $X$ then any closed set belong to $\cal A_\sigma(K)$, what imply that also $\cal T\subset A_\sigma(K)$ (by definition of $\sigma$-algebra). Thus $\mathcal {A_\sigma(T)=B}(X)\subset\cal A_\sigma(A_\sigma(K))=A_\sigma (K)$, as desired.
It’s quite correct; in words: indeed all compact sets are Borel, so the $\sigma$-algebra generated by the compacts is a subset of the Borel $\sigma$-algebra. And as $X$ is $\sigma$-compact, so are all closed subsets, so all closed subsets are in the $\sigma$-algebra generated by the compacts, and hence so is the Borel $\sigma$-algebra.