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