Sherman-Morrison Formula with Matrices
Solution 1:
No, your proof does not work as is. Firstly, the $1$ in the bottom righ should really be a matrix (maybe the identity matrix ?).
The main problem though is that $v^TA^{-1}u$ is not a scalar, it is a matrix and you thus cannot divide so carelessly.
I think what you are looking for is the so-called Woodbbury matrix identity. Wikiedia has a great proof of it at https://en.wikipedia.org/wiki/Woodbury_matrix_identity