Normal operator close in norm to projection, what about the spectrum? Let $T$ be a normal operator on a Hilbert space and suppose $T$ is close in norm to a projection $P$. Can I say that the spectrum of $T$ is contained in small balls around $0$ and $1$?
 A: First of all $\rho(T)=||T||$ because $T$ is normal ($\rho$ is the spectral radius). So by the triangle inequality $\rho(T)\leq1+c$, given $||T-P||\leq c$.
Now the formula in the comment 
$$(T-\lambda I)^{-1}=((T-P)(P-\lambda I)^{-1}+I)^{-1}(P-\lambda I)^{-1}$$
makes sense, using the Von Neumann series, if $(T-P)(P-\lambda I)^{-1}$ is less than $1$ in norm.
If we let $A$ be the closed ball centered at $0$ of radius $1+c$, with two small open balls removed around $0$ and $1$, we have $\sup_A ||(P-\lambda I)^{-1}||<\infty$ by compactness. So if $c$ is small enough we have the result. The smaller the open balls, the smaller the $c$.
A: Even with $P$ just an idempotent and not requiring $T$ to be normal. 
We have 
\begin{align}
\|T-T^2\|&\leq\|T-P\|+\|P-P^2\|+\|P^2-T^2\|\\ \ \\
&=\|T-P\|+\|P^2-T^2\|\\ \ \\
&\leq\|T-P\|+\|P^2-PT\|+\|PT-T^2\|\\ \ \\
&\leq (1+\|P\|+\|T\|)\,\|T-P\|.
\end{align}
Now, using $\rho$ for the spectral radius, 
$$\tag{1}
\rho(T-T^2)\leq\|T-T^2\|\leq k\|T-P\|
$$
for a certain $k$ (we can take $k=1+\|P\|+\|T\|$ as above). Let $r=k\|T-P\|$. Since 
$$
\sigma(T-T^2)=\{\lambda-\lambda^2:\ \lambda\in\sigma(T)\}
$$
we get from $(1)$ that $|\lambda-\lambda^2|\leq r$ for all $\lambda\in\sigma(T)$. Thus, either
$$
|\lambda|\leq \sqrt r,\ \ \ \text{ or } \ \ \ |1-\lambda|\leq\sqrt r.
$$
In other words, if $B_\delta(\mu)$ denotes the ball of radius $\delta$ around $\mu$, then $$\sigma(T)\subset B_{\sqrt r}(0)\cup B_{\sqrt r}(1).$$
