Why call this a spectral projection? Regarding this question,
Why do spectral projections give norm approximations?
I have figured out what is meant by spectral projection, and have thus found the answer as well.  A spectral projection is the image of $x$ under a step/indicator function defined on its spectrum, which is hence an orthogonal projection on some closed subspace. (Here I am using the Bounded Borel functional calculus.)
Okay, it's a projection.  And I said the word "spectrum" in the above paragraph.  But surely it's more "spectral" than that.  Does the image of the so-obtained projection have anything to do with $x$ of a spectral nature? (This is an open-ended question, so please qualify what sort of spectral significance there is.  But some specific interpretations are below):


*

*Suppose $\lambda$ is an eigenvalue of $x$ which hence means that its in the spectrum of $x$.  Does the indicator function of the one point $\lambda$ applied to x via the Borel Functional Calculus project onto the eigenspace associated to it?  By considering $z*\chi_\lambda$ one can see that the image of this projection is a subset of the eigenspace of $\lambda$.  Does the other inclusion hold?

*More generally, if $\lambda$ is only a general element of the spectrum, is the indicator function of it applied to $x$ equal to a projection projecting onto some sort of "spectrally significant space for $\lambda$?"
 A: If $\chi_\lambda$ is the indicator of the single point $\lambda$, then the corresponding spectral projection $\chi_\lambda(A)$ for your normal operator $A$ is indeed the orthogonal projection on the kernel of
$A - \lambda I$, i.e. the eigenspace for $\lambda$.  If $\lambda$ is not an eigenvalue, that means the projection is $0$.
EDIT: Why is the first statement true?  Since $z \chi_\lambda(z) = \lambda \chi_\lambda(z)$ for all complex numbers $z$, $A \chi_\lambda(A) = \lambda \chi_\lambda(A)$ which says the range of $\chi_\lambda(A)$ is contained in the eigenspace.  On the other hand, let $g_n(x) = 
1/(x-\lambda)$ for $1/(n-1) \ge |x - \lambda| > 1/n$ (ignore the $1/(n-1)$ in the case $n=1$).  Then $g_n(x)$ is a bounded Borel function, and $E_n(A) = g_n(A) (A - \lambda I)$ is the spectral projection on $S_n = \{x: 1/(n-1) \ge |x-\lambda| > 1/n\}$.  Now if $\psi$ is an eigenvector of $A$ for eigenvector $\lambda$, $E_n(A) \psi = g_n(A) (A - \lambda I) \psi = 0$.  The disjoint union of the sets $S_n$ for positive integers $n$ being ${\mathbb C} \backslash \{\lambda\}$, countable additivity for the projection-valued measure implies 
$$(I - \chi_\lambda(A)) \psi = \sum_{n=1}^\infty E_n(A) \psi = 0$$
i.e. $\psi = \chi_\lambda(A) \psi$ is in the range of $\chi_\lambda(A)$.
