Checking every open cover by basis elements has a finite subcover for compactness Is my use of notation correct here sir? Proof is correct sir?
Let $\mathcal{B}$ be a basis for a space $X$. Then if every cover of $X$ by basic open sets in $\mathcal{B}$ has a finite subcover $X$ is compact.
Let $\mathcal{B}=\{B_k|k \in K\}$ be a basis for $X$. Let $\{O_\kappa|
\kappa \in \Gamma\}$ be an open cover for $X$. Then $X=\bigcup\limits_{\kappa \in \Gamma}O_\kappa$ and for each $\kappa \in \Gamma$ there is a subset $K_\kappa \subset K$ with $O_\kappa=\bigcup\limits_{k \in   K_\kappa}B_k$. Then $X=\bigcup\limits_{\kappa \in   \Gamma}(\bigcup\limits_{k \in   K_{\kappa}}B_k)$ and so there is a finite subcollection $\{B_{k_1},...,B_{k_n}\}$ of $\mathcal{B}$ covering $X$. For each $B_{k_i}$ select a corresponding $O_{k_i}$ containing it. Thus $\{O_{k_1},...,O_{k_n}\}$ is a finite subcover.
 A: It's fine. But if you're doing index sets anyway: define $\Gamma' = \bigcup \{K_\kappa\mid \kappa \in \gamma\}$ and then
$$X = \bigcup_{x \in \Gamma}\left(\bigcup_{k \in \kappa} B_k \right)= \bigcup \{B_k\mid k \in \Gamma' \}$$
and we get a finite subset $F \subseteq \Gamma'$ and for each $f \in F$ there is some $k(f)\in \Gamma$ so that $f \in K_{k(f)}$, unpacking the definitions. But then $\{U_{k(f)}\mid f \in F\}$ is a finite subcover of $\{O_\kappa \mid \kappa \in \Gamma\}$ etc. BTW: Choosing all $K_\kappa$ at the same time is an orgy of AC of course (nothing wrong with it, topologists do it all the time).
Linguistic/notational sidenote: better choose $\gamma \in \Gamma$ etc. For me $\kappa \in \Gamma$ is "weird" (I did Ancient Greek in high school..). I always match letters to each other that belong together ($i \in I$, $n \in \Bbb N$ etc.) For me it reduces confusion. A collection of index sets $I$ could then be $\mathcal{I}$ or $\mathscr{I}$ etc, still of "type" I. For me $k \in \kappa$ is sort of OK because $K$ is both capital kappa as capital k.
And if you want to reduce notation you could go like this: let $\mathcal{B}$ be the base. Let $\mathcal{O}$ be an open cover of $X$.
Because $\mathcal{B}$ is a base for the topology, $\mathcal{O}$ has a refinement $\mathcal{B}'$ by base elements that is still a cover. And $\mathcal{B}'$ has a finite subcover by assumption so $\mathcal{O}$ has a finite refinement and so $X$ is compact.
