proof of inverse of matrix with small-rank adjustment

We will first prove the formula when $A=I$.

Suppose that $R^{-1}+Y^{T}X$ is invertible. Thus

$$(R^{-1}+Y^{T}X){(R^{-1}+Y^{T}X)}^{-1}=I.$$

and

$$R^{-1}{(R^{-1}+Y^{T}X)}^{-1}+Y^{T}X{(R^{-1}+Y^{T}X)}^{-1}=I.$$

Multiply by $XR$ from the left, and multiply by $Y^T$ from theright, we get

$$X{(R^{-1}+Y^{T}X)}^{-1}{Y}^T+XR{Y}^{T}X{(R^{-1}+Y^{T}X)}^{-1}{Y}^T=XR{Y}^T.$$

The right hand side is equal to $B-I$, while the left hand side canbe factorized as

$$(I+XR{Y}^T)X{(R^{-1}+Y^{T}X)}^{-1}{Y}^T.$$

So,

$$B\cdot (X{(R^{-1}+Y^{T}X)}^{-1}{Y}^T)=B-I.$$

After rearranging, we obtain

$$I=B(I-X{(R^{-1}+Y^{T}X)}^{-1}{Y}^T).$$

Therefore

$${(I+XR{Y}^T)}^{-1}=I-X{(R^{-1}+Y^{T}X)}^{-1}{Y}^T$$(*)

For the general case $B=A+XR{Y}^T$, consider

$$B{A}^{-1}=I+XR{Y}^T{A}^{-1}.$$

We can apply (*) with $Y^T$ replaced by $Y^T{A}^{-1}$.