Let $T$ be a bounded self-adjoint operator. I already showed that the operator $\mathbb{1}_{U} (T)$ is an orthogonal projection, where $\mathbb{1}_{U}$ is the characteristic function of the Borel set $U \subset \mathbb{R}$. (To make sense of this, one can think of $\mathbb{1}_U(T) $ as restricted to the spectrum $\sigma(T)$.)
However, now I wish to show that $\mathbb{1}_{\left\{\lambda \right\}} (T)$ is non-zero if and only if $\lambda$ is an eigenvalue of $T$.
My attempt is this: I think we have the identity $$ (x- \lambda) \mathbb{1}_{\left\{ \lambda \right\}}(x) = 0 $$ for all $x \in U$. Then applying the star homomorphism $\Phi$ to this, we get $$ (T - \lambda) \mathbb{1}_{ \left\{ \lambda \right\} } (T) = 0. $$ Does this prove that $\lambda$ is an eigenvalue of $T$?
Yes. If $1_{\{\lambda\}}(T)\ne0$, there exists nonzero $v\in H$ with $1_{\{\lambda\}}(T)v=v$. Then $$ Tv=T1_{\{\lambda\}}(T)v=\lambda 1_{\{\lambda\}}(T)v=\lambda v, $$ so $\lambda $ is an eigenvalue. Conversely, if $\lambda$ is an eigenvalue then $Tv=\lambda v$ for some nonzero $v$, and so you can deduce that $f(T)v=f(\lambda)v$ for all bounded Borel functions $f$ (start with monomials, then polynomials, then take limits). So $$ 1_{\{\lambda\}}(T)v=1_{\{\lambda\}}(\lambda)v=v. $$