= Solution
First construct the <spectral measure of a normal operator> and prove the open-set assertion needed for all three numbered conclusions. The algebra $A=C^*(I,T)$ generated by $T$ and $T^*$ is commutative because $T$ is a <normal operator>. The <Gelfand transform> identifies $A$ with $C(\Phi_A)$. By <spectral permanence for C-star algebras>,
$$
h:\Phi_A\longrightarrow K=\sigma_{\mathcal B(H)}(T),\qquad h(\chi)=\chi(T)
$$
is onto: its image is the spectrum of $T$ in $A$, which is also its spectrum in $\mathcal B(H)$. It is injective because a character is determined by its values on $T,T^*$ and $I$, and $\chi(T^*)=\overline{\chi(T)}$. Thus $h$ is a <homeomorphism>. Apply part (a) and set
$$
P(B)=E(h^{-1}(B))\quad(B\subseteq K\text{ Borel}).
$$
This is a <projection-valued measure> with $P(K)=I$ and
$$
\boxed{T=\int_Kz\,dP(z).}
$$
The associated <continuous functional calculus> is isometric: $\|u(T)\|=\|u\|_\infty$ for $u\in C(K)$.
If $U\subseteq K$ is nonempty and relatively open, choose $z_0\in U$ and a nonzero continuous $u$ supported inside $U$, for example a small continuous bump about $z_0$ in the metric of $K\subseteq\mathbb C$. If $P(U)=0$, then $u(T)=\int_Ku\,dP=0$, contradicting $\|u(T)\|=\|u\|_\infty>0$. Therefore
$$
\boxed{P(U)\ne0\quad\text{for every nonempty relatively open }U\subseteq K.}
$$
This proves the claim from faithfulness of the representation rather than assuming it as part of the spectral theorem.
Now take an <isolated point> $\lambda\in K$. Its singleton is relatively open, so $P(\{\lambda\})\ne0$. Multiplication of spectral integrals gives
$$
(T-\lambda I)P(\{\lambda\})=0.
$$
Thus its range is contained in the <eigenspace> $\ker(T-\lambda I)$. Conversely, for $v$ in that kernel, the positive <scalar spectral measure> $\mu_v(B)=\langle P(B)v,v\rangle$ satisfies
$$
0=\|(T-\lambda I)v\|^2=\int_K|z-\lambda|^2\,d\mu_v(z).
$$
It follows that $\mu_v(K\setminus\{\lambda\})=0$, and hence $\|(I-P(\{\lambda\}))v\|^2=0$. Consequently
$$
\boxed{\lambda\text{ is an eigenvalue},\qquad P(\{\lambda\})\text{ is the orthogonal projection onto }\ker(\lambda I-T).}
$$
The identification of this projection with the <eigenspace> holds even for a non-isolated $\lambda$; isolation guarantees that it is nonzero.
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