= Solution
The real <Stone-Weierstrass theorem> states that if $X$ is compact Hausdorff and $A\subseteq C(X,\mathbb R)$ is a <subalgebra> containing the constants and separating points, then $A$ is uniformly dense in $C(X,\mathbb R)$. Let $B$ be its uniform closure, still an <algebra over a field>. If $h\in B$ and $M\ge\|h\|_\infty$, scale the permitted <polynomial> approximation of absolute value on $[-1,1]$ to $[-M,M]$. Applying these <polynomials> to $h$ proves $|h|\in B$. Therefore
$$
\max(h,k)=\tfrac12(h+k+|h-k|),\qquad
\min(h,k)=\tfrac12(h+k-|h-k|)
$$
belong to $B$: it is a lattice.
Fix $f\in C(X,\mathbb R)$ and $\varepsilon>0$. For every $x,y\in X$, point separation and an affine adjustment give $g_{x,y}\in A$ with $g_{x,y}(x)=f(x)$ and $g_{x,y}(y)=f(y)$; if $x=y$, a constant suffices. For fixed $x$, the open sets where $g_{x,y}>f-\varepsilon$ cover $X$. Select finitely many and take their maximum $h_x\in B$. Then $h_x>f-\varepsilon$ everywhere and $h_x(x)=f(x)$. By <continuity>, $h_x<f+\varepsilon$ on some neighborhood $V_x$ of $x$. Choose finitely many $V_x$ covering $X$ and take the minimum of the corresponding $h_x$. The result $h\in B$ satisfies $f-\varepsilon<h<f+\varepsilon$ everywhere. Since $B$ is closed and $\varepsilon$ arbitrary, $f\in B$. Thus \b[$\overline A=C(X,\mathbb R)$], proving the theorem.
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