Here are some remarks that may be useful. Consider the homogeneous space $X=G/T$ with the torus acting on the left $T\circlearrowright X$. Then it's easy to see that the $T$ fixed points are exactly the cosets in $W=N(T)/T$. At each fixed point $w\in W$, the tangent space $T_wX$ defines a (complex) representation of $T$, which splits into a direct sum of $2$-(real) dimensional weight spaces $T_wX=\bigoplus_{\alpha\in\Phi^+} L_{w^{-1}(\alpha)}$. Define the $2$-dimensional compact submanifolds $P_{w^{-1}(\alpha)}=\operatorname{exp}(L_{w^{-1}(\alpha)})\subset X$. Perhaps it's also not difficult to see (or at least believe) that $P_{w^{-1}(\alpha)}$ is a $T$-invariant $\mathbb{S}^2$, containing exactly two fixed points. Now you have a graph $\Gamma=\left(W,\left\{P_{w^{-1}(\alpha)}\right\}\right)$.

$\textbf{Claim:}$ The graph $\Gamma$ is connected.

I don't really have an easy proof of this. One idea would be to use the convexity of the moment mapping $\mu\colon X\rightarrow\mathfrak{t}^*$, i.e. the image of the moment map is a convex polytope whose graph is a subgraph of $\Gamma$ (containing all vertices). Then recall that the graph of a convex polytope is always connected...

For each index $w^{-1}(\alpha)$, $\alpha\in \Phi^+$, define $s_{w^{-1}(\alpha)}\in W$ as the element $w^{-1}\cdot w'$ where $P_{w^{-1}(\alpha)}$ is the edge joining $w$ and $w'$. Then from the connectivity of $\Gamma$, it's easy to see that $W$ must be generated by $\mathcal{T}=\left\{s_{w^{-1}(\alpha)}\left|w\in W, \ \alpha\in\Phi^+\right.\right\}$. Of course it only remains to show that each of these elements in $\mathcal{T}$ is a reflection. To see this just remark that $s_{w^{-1}(\alpha)}$ must preserve the (kernel of the) weight $\lambda_{w^{-1}(\alpha)}\colon\mathfrak{t}\rightarrow \mathbb{R}$ defining the weight space $L_{w^{-1}(\alpha)}$. And of course since it fixes a hyperplane, it must be a reflection.

definethe Weyl group $W$ of a root system $R$ to be the subgroup of $\mathrm{Aut}(R)$ generated by the reflections in the roots. (For example, see Helgason's or Humphreys' treatment of root systems.) The result that $W=N(T)/T$ depends on the connectedness of $G$, so it sounds as though you will at least have to go into enough detail to show how connectedness of $G$ enters the picture. $\endgroup$ – Robert Bryant Sep 26 '17 at 11:12