Furstenberg's proof I've been researching into topology and came across this proof, however, I don't fully understand it. If possible could someone break down the proof and explain each point at a time?
 A: His proof can also be found in Proofs from THE BOOK, chapter 1. 
The idea is to do a proof by contradiction. You define subsets $S(a,b)=\{an+b:n\in\mathbb{Z}\}$, for $a,b\in\mathbb{Z}, a\neq 0$ of $\mathbb{Z}$ and define a topology on $\mathbb{Z}$ to which uses these sets as basis elements (check that this works!). From this, one can show that all nonempty open sets are infinite.
The next step of the proof is the tricky part. First, you show that $S(a,b)$ is also closed, just by using the axioms of topology and basic properties of the integers. Now, from the above, all nonempty open sets are infinite. In particular, the set $\{\pm 1\}$ is not infinite, so it cannot be open. This means that $\mathbb{Z}\setminus \{\pm 1\}$ cannot be closed. Now, suppose there are finitely many primes $p_1,\dots, p_n$. We use the fundamental theorem of arithmetic to say every integer which is not $\pm 1$ is an integer multiple of a prime. Therefore $\mathbb{Z}\setminus\{\pm 1\}=\bigcup_1^n S(p_i,0)$. But this is the finite union of closed sets, and hence is closed. But this means $\{\pm 1\}$ is open. Contradiction.
That's how the proof works. Is that what you were looking for? 
A: You define a topology by defining what the open sets are.  Furstenberg defines a topology on  the universe of integers $\Bbb{Z}$ by defining the open sets to be the empty set and every other set $X$ that has the following property:
 * $X$ is open if for every integer $x \in X$ there exists some integer $a \neq 0$ such that all the integers of the form $an+x$ are also in $X$.
For example, the set of all evens is open, since for any even number $e$, all numbers of the form $e+2\cdot n$ are also even.  
It is easy to verify that this definition of "open" satisfies all the requirements for being a topology.
Now notice that in this topology, every open set is infinite, and "basis" sets that contain only one arithmetic sequence $an+b$ for all $n\in \Bbb{Z}$ are both open and closed. (A closed set is the complement of an open set. For basis sets, the complement of the set is the union of all sequences of the form $an+b^\prime$ where $0 \leq b^\prime <a$ and $b^\prime \neq b$. Since any union of open sets is itself open, that union is open, hence its complement,  the basis set based on $b$, is closed.) 
Now Furstenberg observes that the only numbers that are not integer multiples of some prime number are $+1$ and $-1$. Consider the set $K$ of numbers that are integer multiples of prime numbers: 
$$ K = \bigcup_{p \text{ prime}} \left[ np:n\in \Bbb{Z} \right] $$
The complement $K$, $\{ -1, +1\}$ is finite so it can't be open; so $K$ can't be closed.
But $K$ is the union of a bunch of basis sets, one for each prime.  Now the union of a finite number of closed sets is itself closed; but $K$ is not closed, so this must be an infinite union.  Thus the set of primes cannot be finite.
