Number is in spectrum iff spectral projection is not zero I have encountered this exercise in my functional analysis class

Let $A$ be a bounded, self-adjoint operator on a Hilbert space $H$. We are asked to prove that $\lambda \in \sigma(A)$, where $\sigma(A)$ is the spectrum of $A$, if and only iff $P_{(\lambda-\epsilon,\lambda+\epsilon)}(A) \neq 0$ for all $\epsilon \in \mathbb{R}$. $P_{(\lambda-\epsilon,\lambda+\epsilon)}(A)$ denotes a spectral projection of $A$.

I am quite stumped, I do not see any relation between $\lambda$ being in the spectrum and $A$ having a non-zero spectral projection on any symmetric interval of $\lambda$. I thank all helpers who can show me how to solve this.
 A: If it has to hold for all $\epsilon$, it has to hold for $\epsilon$ arbitrarily small. Since $A$ is self-adjoint and bounded, the spectrum is real and closed. Therefore, if $\lambda$ is not in the spectrum, it is a minimum distance away from it and the spectral projection onto an small interval around it is zero.
To make this more explicite: Using the spectral theorem, we can express $A$ through a projection-valued measure $\pi_A$ such that
$$A = \int_{\sigma(A)} \lambda~ d\pi_A(\lambda)$$
Its projection is then given by multiplication with the characteristic function:
$$P_{[a,b]}(A) = \int_{\sigma(A)} \chi_{[a,b]}(\lambda) \lambda ~ d\pi_A(\lambda).$$
From this, it becomes clear that $P_{[a,b]}(A) = 0$ if $[a,b] \cap \sigma(A) = \emptyset$. If $[a,b] \cap \sigma(A) \neq \emptyset$, then $\chi_{[a,b]} d\pi_A(\lambda) \equiv d\pi_{\tilde A}$ is again a measure on $[a,b] \cap \sigma(A)$ with a corresponding operator
$$\tilde A = P_{[a,b]}(A) = \int_{[a,b] \cap \sigma(A)} \lambda d\pi_{\tilde A}(\lambda).$$
Is this more clear (and correct)? Or is there an easier way to prove this?
