 Solution

The real Stone-Weierstrass theorem states that a subalgebra $A\subseteq C(K,\mathbb{R})$ on a compact Hausdorff space $K$, containing constants and separating points, is dense in the uniform norm. Let $B$ be its uniform closure. Polynomial approximation to $|t|$ on a bounded interval shows that $f\in B$ implies $|f|\in B$. Therefore $B$ is closed under pointwise maximum and minimum, since $\max (f,g)=(f+g+|f-g|)/2$ and $\min (f,g)=(f+g-|f-g|)/2$.

Fix $h\in C(K)$ and $ϵ>0$. Point separation and constants supply, for each pair $x,y$, a function $f_{xy}\in A$ agreeing with $h$ at both points. For fixed $x$, finitely many open sets where $f_{xy}>h-ϵ$ cover $K$; their corresponding maximum $g_x\in B$ satisfies $g_x>h-ϵ$ everywhere and $g_x(x)=h(x)$. The open sets where $g_x<h+ϵ$, as $x$ varies, cover $K$. Choose finitely many and take their minimum $g\in B$. Then $h-ϵ<g<h+ϵ$ throughout $K$. Approximating $g$ by members of $A$ proves the theorem.

Now let $F$ be the uniform closure of integer-coefficient polynomials on $[a,b]$. It is a closed ring. Since $q={\max }_{[a,b]}|1-2x|<1$, the integer-coefficient polynomials ${\sum }_{j=0}^{N}x(1-2x{)}^j$ converge uniformly to $1/2$, with error at most $b{q}^{N+1}/(1-q)$. Thus $1/2\in F$. Ring operations give every dyadic rational constant, and closure gives every real constant. Since $x\in F$, all real polynomials lie in $F$. The Stone-Weierstrass theorem now yields $\boxed{F=C([a,b])}$, as in integer polynomial approximation away from zero and one.

On $[0,b]$ the answer is no. The constant function $1/2$ cannot be uniformly approximated by integer-coefficient polynomials: at zero every such polynomial has an integer value and therefore error at least $1/2$.