I'm reading a proof that the map $A\mapsto A^{-1}$ is continuous in the operator norm. A part of the proof is that if $A,B$ are matrices such that $A$ is invertible and $\beta=\parallel B-A\parallel_{op},\alpha=1/\parallel A^{-1}\parallel_{op}, \beta<\alpha$ then $B$ is invertible and $\parallel B^{-1}\parallel_{op}\leq\frac{1}{\alpha-\beta}$. I'm having some hard time with the final conclusion of that part which is: $|B^{-1}x|\leq\frac{1}{\alpha-\beta}|x|\Rightarrow\parallel B^{-1}\parallel_{op}\leq\frac{1}{\alpha-\beta}$ The conclusion seems wrong to me since I can't prove that equality here in the general case: $\forall v\in\mathbb{R}^n, |Av|\leq\parallel{A}\parallel_{op}\cdot|v| $. Granted, I haven't given it much thought but intuitively it should hold for eigenvectors of the largest eigenvalue and there's no reason it should hold otherwise.
I guess my question would be, why is this result correct?