The disc algebra is unital and separating but not self-adjoint or dense

The reverse triangle inequality derived from [L3] makes $z\mapsto \mathrm{\mid }z\mathrm{\mid }$ continuous, so $D=\{\mathrm{\mid }z\mathrm{\mid }\le 1\}$ is closed; it is bounded because $d_{\mathbb{C}}(z,0)=\mathrm{\mid }z\mathrm{\mid }\le 1$. Thus [L5], [L12], and [L13] make $D$ compact, and [L14] makes its metric topology Hausdorff.

Each $p={\sum }_{j=0}^m{\alpha }_j{z}^j$ in $P$ is continuous: the identity $z^j-w^j=(z-w){\sum }_{k=0}^{j-1}{z}^k{w}^{j-1-k}$ of [L6] together with $\mathrm{\mid }zw\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }$ and $\mathrm{\mid }z+w\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }+\mathrm{\mid }w\mathrm{\mid }$ from [L3] gives $\mathrm{\mid }{z}^j-w^j\mathrm{\mid }\le j\mathrm{\mid }z-w\mathrm{\mid }$ for $z,w\in D$, so $\mathrm{\mid }p(z)-p(w)\mathrm{\mid }\le C_p\mathrm{\mid }z-w\mathrm{\mid }$ with $C_p:={\sum }_{j=1}^{m}j\mathrm{\mid }{\alpha }_j\mathrm{\mid }$, which is continuity for the metric of [L5]; hence [L15] puts every member of $A$ in $C(D,\mathbb{C})$. The set $P$ contains the constants and the coordinate function $z\mapsto z$, which separates points, and is closed under complex linear combinations and products by [L1] and [L4], so $P$ is a unital point-separating complex function algebra and $A\supseteq P$ inherits unitality and point separation. $A$ is a complex vector subspace because approximants add and scale. For products, $\mathrm{\mid }z\mathrm{\mid }\le 1$ and [L3] give $\mathrm{\mid }p(z)\mathrm{\mid }\le M_p:={\sum }_{j=0}^m\mathrm{\mid }{\alpha }_j\mathrm{\mid }$ on $D$ for every $p\in P$, so given $f,g\in A$ and $\eta >0$ one may first fix $b_0\in P$ with $\mathrm{\mid }g-b_0\mathrm{\mid }<1$ everywhere, whence $\mathrm{\mid }g\mathrm{\mid }\le K:=M_{b_0}+1$ on $D$, and then choose $a\in P$ with $\mathrm{\mid }f-a\mathrm{\mid }<\eta \mathrm{/}(2K)$ everywhere and $b\in P$ with $\mathrm{\mid }g-b\mathrm{\mid }<\eta \mathrm{/}(2(M_a+1))$ everywhere; from $ab-fg=a(b-g)+(a-f)g$ and [L3], $\mathrm{\mid }ab-fg\mathrm{\mid }\le M_a\mathrm{\mid }b-g\mathrm{\mid }+K\mathrm{\mid }a-f\mathrm{\mid }<\eta$ pointwise, and $ab\in P$, so $fg\in A$. Finally $A$ is uniformly closed, because a function within $\epsilon \mathrm{/}2$ of a member of $A$ everywhere is within $\epsilon$ of a member of $P$ everywhere.

Suppose for contradiction that $\bar{z}\in A$. Then there is a nonzero polynomial $p(z)={\sum }_{j=0}^m{a}_j{z}^j$ with ${\sup }_{z\in D}\mathrm{\mid }\bar{z}-p(z)\mathrm{\mid }<1$; put $N:=m+2\ge 2$ and $\zeta :=\exp (2\pi i\mathrm{/}N)$.

Repeated use of the addition law [L9], along the induction of [L11] on $k$ with base $\exp 0=1={\zeta }^0$, gives $\exp (2\pi ik\mathrm{/}N)={\zeta }^k$ for every natural $k$; so the list of [L7] is exactly $1,\zeta ,\dots ,{\zeta }^{N-1}$, and these are the $N$th roots of unity. For an integer $r$ with $1\le r<N$ one has ${\zeta }^r=\exp (2\pi ir\mathrm{/}N)$, and [L10] makes this equal to $1=\exp 0$ only when $2\pi ir\mathrm{/}N\in 2\pi i\mathbb{Z}$, that is only when $N$ divides $r$, which fails in that range; hence ${\zeta }^r\mathrm{\ne }1$. The same law gives $({\zeta }^r{)}^N=\exp (2\pi ir)=1$.

For every natural $q\ge 1$ and every $x\in \mathbb{C}$, $(x-1){\sum }_{k=0}^{q-1}{x}^k=x^q-1$. Apply [L11] to the property that this identity holds for $q=n+1$. At $n=0$ the identity reads $(x-1)x^0=x-1$, which is immediate. Assuming it at $n$, adding the term $x^{n+1}$ to the sum changes the left side by $(x-1)x^{n+1}=x^{n+2}-x^{n+1}$, carrying the right side from $x^{n+1}-1$ to $x^{n+2}-1$, which is the identity at $n+1$.

The exponent-one cancellation ${\sum }_{k=0}^{N-1}{\zeta }^k=0$ is [L8]. For $2\le r\le m+1<N$, step 1.5 with $x={\zeta }^r$ and step 1.4 give $({\zeta }^r-1){\sum }_{k=0}^{N-1}{\zeta }^{rk}=0$ with ${\zeta }^r-1\mathrm{\ne }0$, so ${\sum }_{k=0}^{N-1}{\zeta }^{rk}=0$.

Every sampled point lies on the unit circle, so [L3] gives ${\zeta }^k\overline{{\zeta }^k}=1$ and hence $N^{-1}{\sum }_{k=0}^{N-1}{\zeta }^k\overline{{\zeta }^k}=1$.

Expanding $p$ and using step 2.1 for the exponents $j+1\in \{1,\dots ,m+1\}$ gives $N^{-1}{\sum }_{k=0}^{N-1}{\zeta }^{k}p({\zeta }^k)=0$.

Subtracting step 3.1 from step 2.2 and repeatedly applying the triangle inequality in [L3], justified over the finite sum by [L11], yields $1\le N^{-1}{\sum }_{k=0}^{N-1}\mathrm{\mid }\overline{{\zeta }^k}-p({\zeta }^k)\mathrm{\mid }<1$, contradicting step 1.3.

Therefore $\bar{z}\mathrm{\notin }A$. Since the coordinate function $z$ belongs to $A$, the algebra is not self-adjoint by [L1]. The conjugation map is continuous, because $\bar{z}-\bar{w}=\overline{z-w}$ by [L2] and $\mathrm{\mid }\bar{u}{\mathrm{\mid }}^2=\bar{u}u=\mathrm{\mid }u{\mathrm{\mid }}^2$ by [L3], so $\mathrm{\mid }\bar{z}-\bar{w}\mathrm{\mid }=\mathrm{\mid }z-w\mathrm{\mid }$; and $A$ is uniformly closed by step 1.2, so a function uniformly approximable by members of $A$ lies in $A$. Hence $\bar{z}$ is a member of $C(D,\mathbb{C})$ that $A$ cannot approximate uniformly, and $A$ is not dense.