distance between sets in a metric space

For 1, $S$ might be unbounded, contrary to your statement. But you have exactly the right idea: it is bounded below by 0, and that is all you need.

$\def\dist{\operatorname{dist}}$But for 2 your idea is no good. Let $A = (-1,0)$ and $B = (0, 1)$. Then $A\cap B = \phi$, and yet $\dist(A, B) = 0$. There are no two points $a\in A, b\in B$ with distance zero, as you said, but there are points $a,b$ at arbitrarily small distance, and that is all we need for the infimum of $S$ to be zero. So if we want to prove that $\dist(A,B)>0$, it is not enough to know that $A$ and $B$ are disjoint; we need some additional information about $A$ and $B$. That is where the compactness comes in.

In fact it's not even enough for the two sets to be closed; the usual example is that the hyperbola $H = \{ \langle x,y\rangle \mid xy = 1 \}$ is closed and the $x$-axis $L = \{ \langle x,y\rangle \mid y = 0 \}$ is closed, but $\dist(H, L) = 0$ because for any positive $s$ we can choose $0<t<s$ and then $h = \left\langle \frac1t, t\right\rangle, \ell = \left\langle \frac1t, 0\right\rangle$ are closer than $s$.