Why call this a spectral projection?

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 $\chi_\lambda(A)$.