First few paragraphs will contain a lot of assumptions and statements so I've marked the start of the exact question below.
Let $H$ be a Hilbert space over $\mathbb{C}$ and $T$ be an essentially self-adjoint unbounded linear operator (*) such that 1.) $T$ has a sequence of eigenvectors $\{e_n\}_n$ which form a basis for $H$ and 2.) if $f:\mathbb{N}\to\mathbb{C}$ gives the eigenvalue of $e_n$, then $0 \leq f(1) \leq f(2)\leq \cdots \to \infty$, i.e. all eigenvalues are real, positive and grow unboundedly.
(*) this means that i.) domain $D(T)$ of $T$ is dense in $H$, ii.) $T$ is symmetric, iii.) closure of the graph $G(T)$ of $T$ is the graph of some self-adjoint operator.
Suppose that $\exists \psi\in H:\exists \lambda\in\mathbb{R}_{> 0}:\|T\psi - \lambda\psi\| \leq C$ for some $C > 0$. Note that $\psi$ is not necessarily an eigenvector of $T$. Denote by $\sigma(T)$ the spectrum of $T$. If $E\subset\mathbb{C}$, define
$$S_E := \mathrm{span}\left(\{\phi\in H:\exists c\in E:T\phi = c\phi\}\right)_\mathbb{C}$$
and denote by $P_E:H\to S_E$ the orthogonal projection onto $S_E$.
My question/issue: Now take $E := [\lambda - 2C, \lambda + 2C]$ for the aforementioned $\lambda, C$. Why does it then follow that $\|P_E\psi\|^2\geq \frac{3}{4}$?
Edit: $S_E$ was fixed to be the appropriate span of vectors.
So, it's necessary to assume that $\|\psi\| = 1$.
We can write $$\psi = \sum_{i=1}^\infty a_i e_i$$ where $$\|\psi\|^2 = \sum_{i=1}^\infty |a_i|^2 = 1$$
Let $A=\{i\in \Bbb{N}^+: f(i)\in E\}$ and $B=\Bbb{N}^+\setminus A$. Then for $i\in B$, $|f(i)-\lambda|> 2C$. So $$ \begin{align} C^2 \ge \|(T-\lambda\mathbf{1})\psi\|^2 &= \sum_{i=1}^\infty |(f(i)-\lambda) a_i|^2\\ &\ge \sum_{i\in B} |(f(i)-\lambda) a_i|^2\\ &\ge 4C^2 \sum_{i\in B} |a_i|^2 \end{align}$$ and therefore $$\sum_{i\in B} |a_i|^2 \le \frac{1}{4}$$ So $$ \begin{align} \|P_E\psi\|^2 &= \sum_{i\in A} |a_i|^2\\ &= \sum_{i=1}^\infty |a_i|^2 - \sum_{i\in B} |a_i|^2\\ &= 1 - \left(\le \frac{1}{4}\right)\\ &\ge \frac{3}{4} \end{align}$$