I have NO idea how to approach this problem. I don't see any connection between the corollary and the formula we need to prove. Does anyone have any hints?
Corrolary: If $\mathbf{A}$ and $\mathbf{B}$ are nonsingular matrices of dimensions $m\times m$ and $n\times n$, respectively, and $\mathbf{C}$ and $\mathbf{D}^\intercal$ are arbitrary $m\times n$ matrices, and $\mathbf{A+CBD}$ is nonsingular, then $$ (\mathbf{A+CBD})^{-1} = \mathbf{A}^{-1} - \mathbf{A}^{-1}\mathbf{C}(\mathbf{B}^{-1}+\mathbf{D}\mathbf{A}^{-1}\mathbf{C})^{-1}\mathbf{DA}^{-1}$$
Use the preceding corollary to prove the following updating formula which is very useful for case deletion diagnostics in regression analysis $$ (\mathbf{X}_{-i}^\intercal\mathbf{X}_{-i})^{-1} = (\mathbf{X}^\intercal\mathbf{X})^{-1} + \dfrac{(\mathbf{X}\intercal\mathbf{X})^{-1}\mathbf{x}_i\mathbf{x}_i^\intercal(\mathbf{X}\intercal\mathbf{X})^{-1}}{1-h_{ii}}$$ where $$\mathbf{X} = \begin{pmatrix} \mathbf{X}_{-i} \\ \mathbf{x}_i^\intercal\end{pmatrix} \qquad h_{ii}=x_i^\intercal(\mathbf{X}\intercal\mathbf{X})^{-1}\mathbf{x}_i$$ You may assume that the inverses of $\mathbf{X}^\intercal\mathbf{X}$ and $\mathbf{X}_{-i}^\intercal\mathbf{X}_{-i}$ exist.
I got it:
$$\mathbf{X}^\intercal\mathbf{X} = \mathbf{X}_{-i}^\intercal\mathbf{X}_{-i} + \mathbf{x}_i\mathbf{x}_i^\intercal$$ Therefore $$\mathbf{X}_{-i}^\intercal\mathbf{X}_{-i} =\mathbf{X}^\intercal\mathbf{X} - \mathbf{x}_i\mathbf{x}_i^\intercal = \mathbf{A}+\mathbf{BCD}$$ where $\mathbf{A} = \mathbf{X}^\intercal\mathbf{X}$, $\mathbf{B} = -\mathbf{x}_i$, $\mathbf{C}=1$, and $\mathbf{D} = \mathbf{x}_i^\intercal$. Plug and chug gives us $$\begin{align*} (\mathbf{X}_{-i}^\intercal\mathbf{X}_{-i})^{-1} &= (\mathbf{X}^\intercal\mathbf{X})^{-1} + (\mathbf{X}^\intercal\mathbf{X})^{-1}\mathbf{x}_i(1-\mathbf{x}_i^\intercal(\mathbf{X}^\intercal\mathbf{X})^{-1}\mathbf{x}_i)^{-1}\mathbf{x}_i^\intercal(\mathbf{X}^\intercal\mathbf{X})^{-1}\\ &= (\mathbf{X}^\intercal\mathbf{X})^{-1} + (\mathbf{X}^\intercal\mathbf{X})^{-1}\mathbf{x}_i(1-h_{ii})^{-1}\mathbf{x}_i^\intercal(\mathbf{X}^\intercal\mathbf{X})^{-1}\\ &= (\mathbf{X}^\intercal\mathbf{X})^{-1} + \dfrac{(\mathbf{X}^\intercal\mathbf{X})^{-1}\mathbf{x}_i\mathbf{x}_i^\intercal(\mathbf{X}^\intercal\mathbf{X})^{-1}}{1-h_{ii}} \end{align*}$$