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

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

By Alexander:

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

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

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

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

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

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

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

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

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