This is only a slight modification of the argument already given, but I like it enough to type it in.

Since $W$ acts with no stabilizer on the open Weyl chamber $K$, for $K$ to contain a root, it cannot be perpendicular to any other root. In particular the Dynkin diagram has to be a complete graph (but the triangle $K_3 = \widehat A_2$ is already infinite type), and by inspection of $B_2,G_2$ we see that every root is perpendicular to some other root. So the only possibilities are $A_1,A_2$.

This is only a slight modification of the argument already given, but I liked it enough to type it in.

Since $W$ acts with no stabilizer on the open Weyl chamber $K$, for $K$ to contain a root $\beta$, this $\beta$ cannot be perpendicular to any other root. In particular the Dynkin diagram $D$ has to be a complete graph (but the triangle $K_3 = \widehat A_2$ is already infinite type, so as a graph $D$ has to be $K_1$ or $K_2$), and by inspection of $B_2,G_2$ we see that every root is perpendicular to some other root. So the only possibilities are $A_1,A_2$.