Why this is the same as Euclid's proof:
The proof works by defining a topology on Z generated by arithmetic sequences S(a,b) = {ak+b; k in Z}. The thing is, to prove these things we have to go into number theory and then come back into topology, so nothing is really gained.
To prove that these arithmetic sequences form a basis for a topology, we need to show that every S(a,b) and S(c,d) and for ever N in the intersection of S(a,b) and S(c,d), there exists some S(x,y) fully contained in the intersection that also contains N. If N is in S(a,b) and S(c,d), and L is any common multiple of a and c, then N is in S(L,N) which is contained in both S(a,b) and S(c,d). This is where Euclid's Argument lies, just written in a more converse language. If M=ac+1 is in the intersection of S(a,b) and S(c,d), then S(ac, M) is totally contained in it as well. In particular, 1 = -ac+M is in the intersection. Using induction and applying it to the primes says that if N=p1...pn+1 is a multiple of the pi (ie in the intersection of the S(pi,0)) then so is 1.
Euclidean Division can be stated in this topology as the fact that S(a,b) is both open and closed. This is because the sets {S(n,r); 0<=r<n} cover all of Z and are disjoint. This is the same as saying that for all N, there is a unique r with 0<=r<n so that N-r is a multiple of n, which is what Euclidean Division states. So to prove that S(a,b) is open and closed, we need to appeal to Euclidean Division.
Euclid's Lemma amounts to the fact that the union of all S(p,0), over all primes p, is all of Z except {-1,1}. That is, for every integer N, not -1 or 1, there is some p that divides N, which means that N is in S(p,0). So we need to appeal to Euclid's Lemma to state that the union of all S(p,0) is everything except {-1,1}
Now that we have actually used every part of Euclid's proof, we can go through a few more steps to prove it in the Topology setting. Since Z is infinite, each S(a,b) is infinite and so a finite set cannot be open and so the compliment of a finite set cannot be a closed set in this topology. So {-1,1} is not closed. Therefore, if there are only finitely many primes, the union of all S(p,0) would be finite and since each S(p,0) is closed, the union would be closed. But then the compliment of this would be finite, which cannot happen. Therefore, there must be infinitely many primes.
Hopefully this demonstrates that Furstenberg's Theorem is just Euclid's Theorem cosplaying as topology. No topological results were used and to be able to state the topological parts, we needed to actually do the number theory.