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 $\chi_\lambda(A)$.
For simplicity consider the multiplication operator $M$ on $L^2[0,1]$.
Define a projection valued measure $\mu$ by $\mu(A) \psi =\chi_{A} \psi$ for $ A \in \mathcal{B}([0,1])$ and $\psi \in L^2[0,1]$.
Let $\varphi , \psi \in L^2$ arbitrary. Then for any $ A \in \mathcal{B}([0,1])$
$$ \mu_\varphi^\psi (A) := \langle \varphi , \mu(A) \psi\rangle = \int_0^1 \chi_{A} (x)\bar{\varphi}(x) \psi(x) d x . $$
And so $ \mu_\varphi^\psi $ has the density $\bar{\varphi} \psi $ with respect to the Lebesgue measure on $[0,1]$.
Therefore
$$
\int_{ [0,1]} \lambda d\mu_\varphi^\psi (\lambda) = \int_0^1 \lambda \bar{\varphi}(\lambda ) \psi(\lambda) d \lambda = \langle \varphi, M \psi \rangle
$$
showing that $\mu$ is the spectral measure of $M$.
For $E \in \mathcal{B}([0,1])$ the spectral subspace $V_E$ is defined by
$$V_E = \mathrm{ran} \, \mu (E) = \{ \psi \in L^2 : \exists \varphi \in L^2 : \psi = \chi_E \varphi \} = \{ \psi \in L^2 : \psi|_{[0,1]\setminus E } = 0 \},$$
which is exactly the expression in your post.
There is a big difference between "generalized eigenvectors" (in the distributional sense, which is what the author is reffering to here) and "approximate eigenvectors" (sequence of vectors in the Hilbert space with certain properties). They are not the same.
Of course no distribution is in the spectral subspace.
The author is merely guessing from the distributional eigenvectors how the spectral subspaces might look. Perhaps this can also be justified by using the nuclear spectral theorem.
Best Answer
Yes. If $\chi_{\{\lambda\}}(A)\ne0$, then any vector in the range of $\chi_{\{\lambda\}}(A)$ is an eigenvector for $\lambda$.
An operator may have no eigenvalues. If all the elements of $\Omega$ happen to be eigenvalues, then indeed $\chi_\Omega(A)$ is the projection onto the closure of the linear span of the union of the eigenspaces.
Edit: an example. Let $H=\mathbb C\oplus L^2[0,1]$. Let $$ A=2\oplus M_g, $$ where $g(t)=t\,1_{[1,3]}(t)$ and $M_g$ is the multiplication operator. Then $$ \chi_{\{\frac12\}}(A)=1\oplus0,\qquad\text{ while }\qquad \chi_{[1,2]}(A)=1\oplus M_h, $$ with $h=1_{[1,3]}$.