Because $x_1,x_2\geqslant0$ we can write the inequalities as equalities with the use of slack variables:
Extreme points are equivalent to basic feasible solutions. Because $S\subset\mathbb R^2$, a basic solution has two basic variables; hence there are six candidates: $$(x_1,x_2), (x_1,s_1), (x_1,s_2), (x_2,s_1), (x_2,s_2), (s_1,s_2).$$
Solve the system of equations by setting each nonbasic variable equal to zero; if the resulting solution is positive, then it is feasible, and hence a basic feasible solution. I'll leave the computations to you, but the extreme points we find in this process are $(0,0), (3,0), (0,2)$ (in terms of $(x_1,x_2)$).