Suppose that $T,U$ are bounded operators on a Hilbert space $H$, with $U$ unitary. If $\|T-U\| < 1$, then show that $T$ is invertible.
Injectivity is easy: Suppose $z \in \ker T$, then $\| (T-U)z \| < \|z\|$ for $z \neq 0$ which implies that $z = 0$. So $T$ has to be injective.
What about surjectivity?
The standard way to show this uses a result from Banach algebra theory (if I can think of a more elementary way to show this, I'll edit my answer to include it).
Let $A$ be a unital Banach algebra with unit $1_A$. If $a\in A$ and $\|a-1_A\|<1$, then $a$ is invertible.
If you haven't seen this, a proof should be found in almost any book on functional analysis, and definitely in one which has a chapter on bounded operators on Banach spaces (or a chapter on Banach algebras).
Returning to the question at hand: since $\|T-U\|<1$, we have $$\|U^*T-1\|=\|U(U^*T-1)\|=\|T-U\|<1.$$ Thus $U^*T$ is invertible, and therefore $T$ is invertible.
Neumann series.
Show that $$\|1-TU^{-1}\|<1$$ and
$$ TU^{-1}\sum_{k=0}^\infty (1-TU^{-1})^k =\text{Id} $$
just how you'd show that
$$ \sum_{k=0}^{\infty}q^k=(1-q)^{-1} $$
