$\displaystyle A$ isn't empty because $\displaystyle x \in A$ (the one Opalg took when he said "fix $\displaystyle x \in S$").

Another way to look at Opalg's proof is this. Assume that S is indeed connected, and towards a contradiction, assume that S is not path connected, ie. there exist points $\displaystyle x,y \in S$ such that there is no path between x and y. Then take the same sets A and B as before - the last assumption means that B is not empty, and as we've seen A is not empty. Then $\displaystyle S = A \cup B$ where A,B are disjoint (by definition), nonempty and both open, so we get a contradiction to S being a connected set.

I see you agree that A is open. Now take some point, say $\displaystyle b \in B$ and look at a ball $\displaystyle B(b,r) \subseteq S$ around b (which exists since S is open). Now,

1) We know that if $\displaystyle c \in B(b,r)$ then you can construct a path from b to c, namely the straight line from b to c.

2) If there isany$\displaystyle a \in B(b,r)$ such that $\displaystyle a \in A$, then by the definition of A there is a path from x to a. But $\displaystyle a \in B(b,r)$, and by (1) there exists a path from a to b. By "gluing" the two paths together, we get a path from x to b. but $\displaystyle b \in B$, which means that there exists no path from x to b - contradiction.

Can you point what step, exactly, is not clear in this proof?