How do I show that the inversion mapping for linear transforms is continuous in the operator norm? I'm working through some analysis textbooks on my own, so I don't want the full answer. I'm only looking for a hint on this problem.
My question is related to this question, but the textbook I'm working through approaches it without (directly) using matrices. The textbook I'm using states that "when using the operator norm as a metric, the inversion map is continuous" but I'm struggling to prove that.
I know I need to prove that given that $S$ is an invertible linear transform from $\mathbb{R}^n$ to $\mathbb{R}^n$, if I pick some $\epsilon > 0$, there is a $\delta > 0$ such that $||(T+S)^{-1} - S^{-1}|| < \epsilon$ if $||T|| < \delta$ and $T$ is some linear transform between the same spaces.
That's the definition of continuity in the operator norm.
$T$ doesn't necessarily have to be invertible, because (for example) if $S$ is the linear transform represented by the $n \times n$ identity matrix, $T$ could be the matrix of all zeros (the zero transform), and obviously $S+T$ is still invertible, but $T = 0$ isn't.
I understood the book's proof of the lemma that for any invertible linear transform $S$, there is some $\delta > 0$ and $M > 0$ such that if $||T|| < \delta$, then $||(S+T)^{-1}|| < M$ but that's the only lemma that's given in this section. 
Is this all I (somehow) need to prove continuity, along with the definitions of the operator norm, or is their a trick I'm missing?

Here are some algebraic manipulations that may serve as a hint for you. 
$$
\|(T+S)^{-1}-S^{-1}\|=\frac{\|(T+S)^{-1}S-S^{-1}S\|}{\|S\|}=\frac{\|(T+S)^{-1}T\|}{\|S\|}=\frac{\|(T+S)^{-1}\|}{\|S\|}\|T\|
$$
From here you should be able to deduce the result using the information given to you.
