= Solution
For the unheaded representation request, retain the question's standing unital hypothesis. Let $K$ be the <character space of an algebra> of $A$, with its weak-star or <Gelfand topology>. The standard commutative Banach-algebra results allowed here give a nonempty compact Hausdorff <character space> and $\sigma_A(a)=\{\varphi(a):\varphi\in K\}$. In particular, character <compactness> follows from the <Banach-Alaoglu theorem>: the equations $\varphi(1)=1$ and $\varphi(ab)=\varphi(a)\varphi(b)$ define a weak-star closed subset of the dual unit ball.
The <Gelfand transform> is the unital homomorphism
$$
\Gamma:A\to C(K),\qquad \Gamma(a)(\varphi)=\varphi(a).
$$
Every element of the commutative <C-star algebra> is normal. Part i and the character <spectrum> formula therefore give
$$
\|\Gamma(a)\|_\infty=\max_{\varphi\in K}|\varphi(a)|=r(a)=\|a\|.
$$
Thus $\Gamma$ is an isometry, hence injective, and its range is closed because $A$ is complete. Part ii gives $\Gamma(a^*)=\overline{\Gamma(a)}$. The range contains constants and separates points of $K$, since two distinct characters differ on some element of $A$. The complex <Stone-Weierstrass theorem> states that a unital, conjugation-closed subalgebra of $C(K)$ separating points of a compact <Hausdorff space> is uniformly dense. Its application here, together with closedness of the range, yields
$$
\boxed{\Gamma:A\overset{\cong}{\longrightarrow}C(K)\text{ is an isometric }*\text{-isomorphism}.}
$$
This proves the <Commutative Gelfand--Naimark theorem> from the preceding parts and the stated approximation theorem.
The word “every” in the printed deduction must be read under the initial unital hypothesis. If nonunital commutative <C-star algebras> are included, the compact-space statement is false: the nonzero algebra $c_0$ has no multiplicative identity, whereas $C(K)$ for nonempty compact $K$ does. The general nonunital version is $C_0(K)$ on a locally compact <Hausdorff space>, with vanishing at infinity. This qualification preserves the proved unital result.
For the final <real spectrum criterion for a normal C-star element>, the ambient algebra may be noncommutative. Let $B$ be the norm-closed unital subalgebra generated by $x$ and $x^*$. Normality makes these two generators commute. The <involution> preserves their <polynomial> algebra and is continuous, so $B$ is a commutative <C-star subalgebra>.
We first show that $x$ has the same <spectrum> in $B$ as in $A$, without assuming spectral permanence. Put $\Omega=\mathbb C\setminus\sigma_A(x)$. A compact subset of the real axis has connected complement: upper and lower half-plane paths join around either end of a bounded interval containing it, and each omitted real point can be joined locally to the upper half-plane. Define
$$
V=\{\lambda\in\Omega:(\lambda1-x)^{-1}\in B\}.
$$
This set is nonempty because the <Neumann series> puts the inverse in $B$ for $|\lambda|>\|x\|$. It is relatively open: from an inverse in $B$, the local inverse formula
$$
R(\lambda+h)=R(\lambda)\sum_{n\ge0}[-hR(\lambda)]^n
$$
stays in $B$ for small $h$. It is relatively closed by continuity of the ambient resolvent and norm-closedness of $B$. Connectedness gives $V=\Omega$. Conversely an inverse in $B$ is an inverse in $A$, so
$$
\sigma_B(x)=\sigma_A(x)\subset\mathbb R.
$$
This is the connected-resolvent case of <spectrum in a closed unital subalgebra>.
Apply the just-proved commutative C-star representation to $B$. For every character $\psi$ of $B$, $\psi(x)$ lies in this real <spectrum>. Part ii makes $\psi(x^*)=\overline{\psi(x)}=\psi(x)$. The <Gelfand transform> of $B$ is injective by its isometry, so
$$
\boxed{x=x^*.}
$$
For a nonunital ambient <C-star algebra>, perform the same argument in its standard <C-star unitization>, using the usual <unitization> <spectrum>.
Normality cannot be dropped. In $M_2(\mathbb C)$, take
$$
N=\begin{pmatrix}0&1\\0&0\end{pmatrix}.
$$
Since $N^2=0$, its <spectrum> is $\{0\}$: for $\lambda\ne0$, $(\lambda I-N)^{-1}=\lambda^{-1}I+\lambda^{-2}N$, while $N$ is not invertible. Yet $N\ne N^*$ and
$$
NN^*=\begin{pmatrix}1&0\\0&0\end{pmatrix}\ne
\begin{pmatrix}0&0\\0&1\end{pmatrix}=N^*N.
$$
Thus \b[a real <spectrum> alone does not force self-adjointness; normality is essential].
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