= 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 \b[$I$ 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 $\Phi(f,\lambda)=f+\lambda\mathbf1$. Its inverse is $h\mapsto(h-h(0),h(0))$. The specified multiplication and <involution> are preserved by direct substitution. The pullback <norm> $\|(f,\lambda)\|=\|f+\lambda\mathbf1\|_\infty$ is a C-star <norm>, so equals the unique <unitization> <norm> supplied in the question. Therefore \b[$\Phi$ 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 $\|h-p\|_\infty<\varepsilon/2$. Then $b(t)=p(t^2)-p(0)$ is a <polynomial> in the positive even powers of $t$, with zero constant term, and
$$
\boxed{\|a-b\|_\infty\leq\|h-p\|_\infty+|p(0)|<\varepsilon.}
$$
This gives exactly the required finite form.
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