Invertible operators [closed]

I'm doing some homework about functional analysis. I've been using a implication that seemed rather trivial but now that i properly think about it i can't actually prove it. The implication goes

$$``\text{If $T$ isn't invertible then clearly $I-T$ isn't}"$$

In my head this makes sense because $(I-T)x=x-Tx$ and if i can't "invert" $Tx$ i can't invert $x-Tx$.

How could one prove such statment? Is it even correct?

The claim is not true. For example, take $T=0$. Then $I-T$ is invertible.

More generally, $I-T$ is invertible if $\|T\|<1$. So for example, $T=\alpha P$, with $P$ a projection (onto a proper subspace) and $\alpha$ small would be non-invertible but would satisfy $\|T\|<1$.

Almost orthogonally to the above, if $K$ is a compact operator in a Banach space, then there is only a countable many complex scalars $\alpha$ such that $I-\alpha K$ is non-invertible, meaning that $I-\alpha K$ would be invertible “most of the time.”

Understanding small as well as compact perturbations is a fascinating topic that leads to the Riesz-Schauder theory of compact operators, Fredholm alternative, the Fredholm index theory, et cetera.