 Solution

Use the source's notation for the vanishing-at-zero ideal $I=\{f\in C[0,1]:f(0)=0\}$. It is the kernel of bounded evaluation at zero, hence closed. It is stable under pointwise multiplication and complex conjugation, so inherits the C-star identity and completeness. If it had an identity $e$, multiplication against $f(t)=t$ would give $e(t)=1$ for $t>0$, contradicting continuity and $e(0)=0$. Thus $I$ is a non-unital C-star subalgebra. This explicit meaning of the notation matters because the usual notation $C_0$ for a compact space has a different convention.

Define the unitization map by $\Phi (f,\lambda )=f+\lambda 1$. Its inverse is $h\mapsto (h-h(0),h(0))$. The specified multiplication and involution are preserved by direct substitution. The pullback norm $\Vert (f,\lambda )\Vert =\Vert f+\lambda 1{\Vert }_{\infty }$ is a C-star norm, so equals the unique unitization norm supplied in the question. Therefore $\Phi$ is a unital isometric C-star isomorphism onto $C[0,1]$.

For the approximation put $h(s)=a(\sqrt{s})$ on $[0,1]$, so $h$ is continuous and $h(0)=0$. The Stone-Weierstrass theorem supplies a polynomial $p(s)$ with $\Vert h-p{\Vert }_{\infty }<\epsilon /2$. Then $b(t)=p(t^2)-p(0)$ is a polynomial in the positive even powers of $t$, with zero constant term, andThis gives exactly the required finite form.