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 $\mathfrak b$.

The proof depends on how you're setting things up. In my opinion the cleanest approach is the Lie algebraic 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 $\Phi^+$: $\mathfrak b = \mathfrak t \oplus \bigoplus_{\alpha \in \Phi^+} \mathfrak g_\alpha$. The action of $w \in W$ takes $\mathfrak b$ to $\mathfrak b_w = \mathfrak t \oplus \bigoplus_{\alpha \in \Phi^+} \mathfrak g_{w\alpha}$. With respect to the length function defined using $\Delta$, the longest element $w_0$ of $W$ takes $\Phi^+$ to $-\Phi^+$. It follows that $b_{w_0}$ is the Borel subalgebra opposite to $\mathfrak b$.