= Solution
For the real lattice version of the <Stone-Weierstrass theorem>, let $f\in C_{\mathbb R}(\Omega)$ and $\varepsilon>0$. For any $x,y\in\Omega$, separation of points and constants give $h_{xy}\in L$ matching $f$ at both $x$ and $y$: take an affine combination of a separating function, and for $x=y$ take a constant. Fix $x$. The open sets where $h_{xy}>f-\varepsilon$ cover $\Omega$ as $y$ varies. Compactness gives finitely many such functions; their maximum $g_x$ belongs to the lattice $L$, satisfies $g_x>f-\varepsilon$ everywhere, and has $g_x(x)=f(x)$. The open sets where $g_x<f+\varepsilon$ now cover $\Omega$ as $x$ varies. Choose finitely many and take their minimum $h\in L$. Then $f-\varepsilon<h<f+\varepsilon$ everywhere. Thus \b[$L$ is uniformly dense] in the real continuous functions.
For the complex \*-algebra, its real subalgebra $A_{\mathbb R}$ of <Hermitian C-star elements> contains the real and imaginary parts of every member. It separates points, because at least one of those parts distinguishes any pair distinguished by a complex function. Its uniform <closure> is a unital real subalgebra and is a lattice by the allowed result. The real theorem makes this <closure> all of $C_{\mathbb R}(\Omega)$. Approximating the real and imaginary parts separately proves \b[$A$ is uniformly dense in $C(\Omega)$].
On $[0,1]$, the algebra of polynomials in $t^2$ is unital, closed under conjugation, and separates points because $t\mapsto t^2$ is injective there. Choose $p(t^2)$ within $\varepsilon/3$ of $a(t)$, and correct its endpoints by
$$
b(t)=p(t^2)+[a(0)-p(0)](1-t^2)+[a(1)-p(1)]t^2.
$$
This is still a <polynomial> in $t^2$ and has exactly the requested endpoint values. The correction has absolute value at most $\varepsilon/3$, because it is a <convex combination> of two errors of that size. Therefore
$$
\boxed{b(0)=a(0),\quad b(1)=a(1),\quad\|a-b\|_\infty<2\varepsilon/3<\varepsilon.}
$$
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