Verified answer

First, why K is a G-delta. Well, K is closed in X x B'. Now, B' is a G-delta subset of the completely metrizable Y. A standard result in this area (due I think to Alexandroff) implies that B' (with its subspace topology) is therefore a completely metrizable topological space in its own right.

Since X and B' are completely metrizable, so is their product (this is easy: from complete metrics giving the topologies on X and B', it's easy to write one down that gives the product topology on X x B'). So K is closed in X x B', and X x B' is completely metrizable. From a standard result (that you asked in another question) it follows that K is a G-delta.

If you are focusing on details here, K is a G-delta in the topological space X x B' (ie, there is a countable family G_n of subsets of X x B', that are open *in the topology of X x B'* with K = the intersection of the G_n). This implies that K is a G-delta in X x Y but it takes a moment's thought to see this. [The details are simple but nontrivial. The topology on X x B' is the same whether you regard it as a subspace of the product space X x Y, with the subspace topology, or if you instead regard it as the product of the spaces X and B', with B' having the subspace topology from Y. So to say that G_n is open in X x B' means that there is a set U_n open in X x Y with G_n = U_n intersect (X x B'). Since B' is a G-delta in Y, it is clear that X x B' is a G-delta in X x Y (namely: if B' is the intersection of open sets V_n, then X x B' is the intersection of the open sets X x V_n). So the equation G_n = U_n intersect (X x B') exhibits G_n as an intersection of an open subset of X x Y with a G-delta subset of X x Y, and hence we see that each G_n is G-delta in X x Y. So K, a countable intersection of G-delta sets in X x Y, is also a G-delta set in X x Y.

As for the second part, once you know that K is a G-delta set it is pretty simple. Let h denote the function from A' to the product space X x Y given by sending x to the pair (x, f(x)). Since f : A' to Y is continuous it follows (from a short argument) that h: A' to X x Y is also continuous. The set A* is precisely h^{-1}(K) (the inverse image of the set K under the map h). And it is generally true (and easy to check straight from the definitions) that the inverse image of a G-delta set under a continuous map is a G-delta set. So that is why A* is a G-delta set.