I am reading the proof from General Topology by John L. Kelley using Alexanders Subbasis Theorem. I have a few question which I will denote in bold.

Let $Q = \prod_{a \in A} X_a$ where each $X_a$ is compact and $Q$ has the product topology.

Let $\mathcal{S}$ be the subbasis for the product topology consisting of all sets of the form $\pi^{-1}_{a}(U)$, where $\pi_{a}$ is the projection to the a-th coordinate and $U$ is open in $X_a$

By Alexander:

$Q$ is compact if every subcollection $\mathcal{K}$ of $\mathcal{S}$ s.t. no finite subcollection of $\mathcal{K}$ covers $Q$, does not cover $Q$.

For each $a \in A$, let $\mathcal{B}_{a}$ be the collection of all open sets $U$ in $X_a$ s.t. $\pi_{a}^{-1}(U) \in \mathcal{K}$

(Is $\mathcal{B}_{a}$ an open cover of $X_a$?)

Then no finite subcollection of $\mathcal{B}_{a}$ covers $X_{a}$

(is this by hypothesis? If $\mathcal{B}_{a}$ is an open cover of $X_a$, doesn't this contradict the fact $X_a$ is compact?)

then by compactness of $X_a$, $\;\;\;\;\exists$ a point $x_a$ $\in$ $X_a \setminus U$ for all $U$ in $\mathcal{B}_{a}$

(Why does compactness guarantee this point $x_a$?)

The point $x \in Q$ whose a-th coordinate is $x_a$ belongs to no member of $\mathcal{K}$

$\mathcal{K}$ is not a cover of $Q$.
Thank you for your help. I hope I worded my questions clearly enough