Weierstrass Approximation Theorem/Proof 3

Proof
Let $\AA \subseteq \map C {\Bbb I, \R}$ be the set of real polynomial functions.

$\AA$ is a subalgebra of $\map C {\Bbb I, \R}$ because polynomials over $\R$ form an algebra over $\R$.

Let $I$ denote the identity mapping on $\Bbb I$, i.e.:
 * $\forall x \in \Bbb I : \map I x = x$

Then $I \in \AA$.

Thus $\AA$ separates the points of $\Bbb I$, since trivially:
 * $\forall x,y \in \Bbb I : x \ne y \implies \map I x \ne \map I y$

It is also clear that $1 \in \AA$.

Therefore the claim follows from Stone-Weierstrass Theorem.