Suppose $N$ is a normal operator on some $\mathbb{C}$-Hilbert space $\mathcal{H}$ and let $\mathrm{E}:\mathcal{B}(\sigma(N))\to B(\mathcal{H})$ be the associated spectral measure.
Fix $\lambda_0\in\sigma(N)$ and suppose $x\in\mathrm{E}(\{\lambda_0\})=\mathrm{im}(\mathrm{E}(\{\lambda_0\}))$. How would one show that $Nx=\lambda x_0$?
I realize that I should use $\mathrm{E}(\{ \lambda_0\} )=\mathbf{1}_{\{\lambda_0\}}(N)$, but don't know how precisely to apply it. Is projection property the key?
If $x$ is zero the equality holds trivially. If $x\ne0$, now you know that $E(\{\lambda_0\})\ne0$. You have the equality of functions $t\,1_{\{\lambda_0\}}(t)=\lambda_0\,1_{\{\lambda_0\}}(t)$. Applying the bounded Borel functions $t\,1_{\{\lambda_0\}}(t)$ and $\lambda_0\,1_{\{\lambda_0\}}(t)$ to $N$, we get $$ N\,E(\{\lambda_0\})=\lambda_0\,E(\{\lambda_0\}). $$ If you now you apply these two equal operators to $x$, and since $\,E(\{\lambda_0\})x=x$, $$ Nx=\lambda_0\, x. $$