= Solution
The definitions in a unital <C-star algebra> are
$$
\boxed{\begin{array}{ll}
\text{hermitian:}&x^*=x,\\
\text{unitary:}&x^*x=xx^*=1,\\
\text{normal:}&x^*x=xx^*.
\end{array}}
$$
Thus a <Hermitian element of a C-star algebra> is self-adjoint, a <Unitary element of a C-star algebra> has inverse $x^*$, and a <Normal element of a C-star algebra> commutes with its adjoint. We derive the needed <C-star algebra> facts directly from $\|a^*a\|=\|a\|^2$, as required.
First $\|1\|=\|1\|^2$ gives $\|1\|=1$, and
$$
\|a\|^2=\|a^*a\|\le\|a^*\|\|a\|
$$
gives $\|a\|\le\|a^*\|$; applying the same argument to $a^*$ gives equality. Thus the involution is an <isometry>. For a <Unitary element of a C-star algebra> $u$,
$$
\|u\|^2=\|u^*u\|=1,\qquad\|u^{-1}\|=\|u^*\|=1.
$$
The general <Banach algebra> <spectrum> bound gives $|\lambda|\le1$ for $\lambda\in\sigma(u)$. Such a $\lambda$ is nonzero, and the inverse <spectral mapping theorem> gives $\lambda^{-1}\in\sigma(u^{-1})$, so also $|\lambda|^{-1}\le1$. Hence
$$
\boxed{\sigma(u)\subseteq\mathbb T=\{z\in\mathbb C:|z|=1\}.}
$$
If $h=h^*$, the convergent <exponential series> and the isometric involution give $(e^{ith})^*=e^{-ith}$. These commuting exponentials multiply to $1$, so $e^{ith}$ is unitary for real $t$. By the exponential <spectral mapping theorem>, if $\lambda\in\sigma(h)$ then $e^{it\lambda}\in\sigma(e^{ith})\subseteq\mathbb T$. Taking $t=1$ gives $e^{-\operatorname{Im}\lambda}=1$, so
$$
\boxed{\sigma(h)\subseteq\mathbb R.}
$$
Only general <Banach algebra> <spectral mapping theorem> and the defining <C-star identity> were used.
Now prove <spectral permanence for C-star algebras>. Let $B\subseteq A$ be the norm-closed <C-star subalgebra> with the same identity. The algebraic inclusion always gives $\sigma_A(b)\subseteq\sigma_B(b)$. For $h=h^*\in B$, both <spectra> lie in $\mathbb R$. If $\lambda\notin\sigma_A(h)$, choose nonreal $\lambda_n\to\lambda$. Since $\sigma_B(h)\subseteq\mathbb R$, each $(\lambda_n1-h)^{-1}$ belongs to $B$. By <continuity of inversion in a Banach algebra>, these inverses converge in $A$ to $(\lambda1-h)^{-1}$, which belongs to $B$ because $B$ is closed. Thus $\lambda\notin\sigma_B(h)$, proving equality for hermitian elements.
For the requested normal $x\in B$, suppose $b=x-\lambda1$ is invertible in $A$. The element $b^*b$ is hermitian and invertible in $A$, so the equality just proved places $(b^*b)^{-1}$ in $B$. Consequently
$$
y=(b^*b)^{-1}b^*\in B,\qquad yb=1.
$$
Since $b$ already has a two-sided inverse in $A$, multiplying $yb=1$ by that inverse gives $y=b^{-1}$. Hence its inverse belongs to $B$, and
$$
\boxed{\sigma_B(x)=\sigma_A(x).}
$$
We need one more elementary consequence of the <C-star identity>: the <spectral radius norm equality for normal elements>. If $h=h^*$, then $\|h^2\|=\|h\|^2$. If $a$ is normal, commutativity of $a,a^*$ gives
$$
\|a^2\|^2=\|(a^*)^2a^2\|=\|(a^*a)^2\|=\|a^*a\|^2=\|a\|^4.
$$
Every power of $a$ is a <Normal element of a C-star algebra>. Induction gives $\|a^{2^n}\|=\|a\|^{2^n}$. The general <spectral radius formula> therefore implies
$$
r(a)=\lim_{m\to\infty}\|a^m\|^{1/m}=\|a\|.
$$
\b[The continuous functional calculus is the unique unital $*$-homomorphism $C(K)\to\mathcal B(H)$ taking the coordinate function to $T$.] For <bounded linear operators> on a <Hilbert space>, the adjoint identity gives $\|S\|=\sup_{\|v\|=\|w\|=1}|\langle Sv,w\rangle|=\|S^*\|$. Therefore
$$
\|S\|^2=\sup_{\|v\|=1}\langle S^*Sv,v\rangle\le\|S^*S\|\le\|S^*\|\|S\|=\|S\|^2.
$$
This proves the <C-star identity> for $\mathcal B(H)$ directly; completeness and submultiplicativity come from the operator norm. Let
$$
C=\overline{\{p(T,T^*):p\text{ a complex polynomial in two commuting variables}\}}^{\,\|\cdot\|}.
$$
Because $T$ is a <normal operator>, this is a commutative unital <C-star algebra>. Every element of $C$ is a <Normal element of a C-star algebra>, so the preceding norm identity says its <Gelfand transform> is isometric:
$$
\|\widehat a\|_\infty=\sup_{\chi\in\Delta(C)}|\chi(a)|=r_C(a)=\|a\|.
$$
Here $\Delta(C)$ is the compact <character space> from the general <Gelfand representation theorem> for commutative <Banach algebras>.
Each <algebra character> preserves the involution. Indeed, write $a=h+ik$ with $h,k$ hermitian; $\chi(h)\in\sigma_C(h)\subseteq\mathbb R$ and similarly for $k$, so $\chi(a^*)=\overline{\chi(a)}$. Therefore the continuous map
$$
\Delta(C)\longrightarrow K,\qquad\chi\longmapsto\chi(T)
$$
is injective: its value determines $\chi(T^*)$ and hence its value on all the dense <polynomials>. It is surjective because the general character description of the <spectrum> gives $\{\chi(T):\chi\in\Delta(C)\}=\sigma_C(T)$, and <spectral permanence for C-star algebras> identifies this with $\sigma_{\mathcal B(H)}(T)=K$. A <continuous function> that is a bijection from a <compact space> to a <Hausdorff space> is a <homeomorphism>, so we identify $\Delta(C)$ with $K$.
Under this identification, the <Gelfand transform> sends $T$ to $u(z)=z$ and $T^*$ to $\overline u$. Its image is an isometric, and therefore closed, unital self-adjoint subalgebra of $C(K)$. It separates points because it contains $u$. The complex <Stone-Weierstrass theorem> makes the image dense, hence equal to all of $C(K)$. Inverting the <Gelfand transform> and including $C$ into $\mathcal B(H)$ gives
$$
\Phi:C(K)\longrightarrow\mathcal B(H),\qquad f\longmapsto f(T),
$$
an isometric unital <C-star homomorphism> with $u(T)=T$.
To prove uniqueness without assuming automatic continuity of <C-star homomorphisms>, let $\Psi$ be any other such map. Since every $f\in C(K)$ is normal, $\Psi(f)$ is normal. A unital algebra homomorphism preserves inverses, so
$$
\sigma_{\mathcal B(H)}(\Psi(f))\subseteq\sigma_{C(K)}(f)=f(K).
$$
Using the normal-element norm identity yields $\|\Psi(f)\|=r(\Psi(f))\le\|f\|_\infty$. Thus $\Psi$ is contractive. It agrees with $\Phi$ on <polynomials> in $u,\overline u$, since both send them to the same <polynomials> in $T,T^*$. These <polynomials> are dense by the <Stone-Weierstrass theorem>, so continuity proves $\Psi=\Phi$. This establishes the <continuous functional calculus> with no unproved theorem specific to <C-star algebras>.
\b[A disconnected spectrum gives a nontrivial closed invariant subspace.] Write $K=K_1\sqcup K_2$ with $K_1,K_2$ nonempty and both open and closed in $K$. The <indicator function> $f=\mathbf1_{K_1}$ is continuous on $K$, even though no such continuity is needed across the gap outside $K$. Let $P=f(T)$. The <continuous functional calculus> gives
$$
P^2=P,\qquad P^*=P,\qquad PT=TP.
$$
Its isometry gives $\|P\|=1$ and $\|I-P\|=1$, so $P\ne0,I$. The <linear projection> has closed range $Y=PH=\ker(I-P)$. Its range is nonzero and proper, and $T(Px)=P(Tx)\in Y$ proves invariance. Since it also commutes with $T^*$, the same subspace is reducing. This proves <disconnected spectrum gives a reducing subspace>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-106-disconnected-spectrum.png]
{title=A disconnected spectrum and its continuous indicator, which produces a nontrivial reducing projection}
The figure can be realized without any <eigenvalues>: take $H=L^2(K,dA)$, where $dA$ is planar area on the two closed disks, and let $T$ multiply by $z$. Its adjoint multiplies by $\overline z$, so it is a <normal operator>. An <eigenvector> for $\lambda$ would be supported on the area-zero singleton $\{\lambda\}$, hence would be zero in $L^2$. Outside $K$, multiplication by $1/(z-\lambda)$ is a bounded inverse to $T-\lambda I$. For $\lambda\in K$, normalized <indicator functions> of $K\cap\{|z-\lambda|<\varepsilon\}$ have $\|(T-\lambda I)h\|\le\varepsilon$, excluding a bounded inverse. Thus its <spectrum> is exactly $K$. The <linear projection> in the figure multiplies by $\mathbf1_{K_1}$, and its range consists of functions supported on the left disk.
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