The proof depends on how you're setting things up. In my opinion the cleanest approach is pretty clear if you work with the Lie algebrasalgebraic one, and it goes as follows. Your Borel subalgebra $\mathfrak b$ determines a choice of simple roots $\Delta$ and consequently a choice of positive roots $\Phi^+$: $\mathfrak b = \mathfrak t \oplus \bigoplus_{\alpha > 0} \in \Phi^+} \mathfrak g_\alpha$. The action of $w \in W$ takes $\mathfrak b$ to $\mathfrak b_w = \mathfrak t \oplus \bigoplus_{\alpha > 0} \in \Phi^+} \mathfrak g_{w\alpha}$. Since With respect to the length function defined using $\Delta$, the longest element $w_0$ of $W$ takes the positive roots $\Phi^+$ to the negative roots, it $-\Phi^+$. It follows that $b_{w_0}$ is the Borel subalgebra opposite to $\mathfrak b$.
This is pretty clear if you work with Lie algebras. Your Borel subalgebra $\mathfrak b$ determines a choice of positive roots: $\mathfrak b = \mathfrak t \oplus \bigoplus_{\alpha > 0} \mathfrak g_\alpha$. The action of $w \in W$ takes $\mathfrak b$ to $\mathfrak b_w = \mathfrak t \oplus \bigoplus_{\alpha > 0} \mathfrak g_{w\alpha}$. Since $w_0$ takes the positive roots to the negative roots, it follows that $b_{w_0}$ is the Borel subalgebra opposite to $\mathfrak b$.