Applying the Stone-Weierstrass Theorem to approximate even functions 
Let $f:[-1,1] \rightarrow \mathbb{R}$ be any even continuous function on $[-1,1]$ (i.e. $f(-x)=f(x)$ $\forall x \in [-1,1]$). Let $\epsilon>0$. Prove that there exists an even polynomial $p$ such that $$|f(x)-p(x)|< \epsilon$$ $$\forall x \in [-1,1]$$

Here, "even polynomial" means that $p(-x)=p(x)$, not simply that it has even degree.
I think I should use the Stone-Weierstrass theorem to show that the subalgebra of even polynomials, call it $\mathcal{A}$, over this interval is dense, from which the result follows immediately.
For this to work I require that $\mathcal{A}$ contains the constants (obviously true) and separates points...which is not true, unfortunately. Anyone have any hints? I would prefer hints only, rather than solutions.
Oh yes, and I should mention that the version of the Stone-Weierstrass theorem that I can use says that if a subalgebra of $C(\mathbb{R})$ contains the constants and separates points, then it is dense in $C(\mathbb{R})$.
 A: Hint: If $\sup_{x \in [-1,1]} |\,p(x) - f(x)|<\epsilon$ then
 $\sup_{x \in [-1,1]} |\,\frac{1}{2}(\,p(x)+ p(-x)\,)- f(x)\,|<\epsilon$
$\bf{Added:}$
Let $p(x)$ a polynomial so that for every $x \in [-1,1]$ we have $|\,p(x) - f(x)|<\epsilon$. Note that if $x \in [-1,1]$ then also $-x \in [-1,1]$. So we have
$$|\,p(x) - f(x)|<\epsilon\\
|\,p(-x) - f(-x)|<\epsilon$$
Add up the inequality and divide by $2$:
$$\frac{1}{2} ( |\,p(x) - f(x)| + |\,p(-x) - f(-x)|) < \epsilon$$
Note that we have $|a+b| \le |a|+|b|$ for all numbers. Therefore we get 
$$\frac{1}{2} \cdot |(p(x) + p(-x) ) - (f(x) + f(-x) ) | < \epsilon$$
or
$$ |\frac{1}{2}\cdot (p(x) + p(-x) ) - \frac{1}{2}\cdot(f(x) + f(-x) ) | < \epsilon$$
$\tiny{\text{ the averages also satisfy the inequality }}$. 
Now since $f$ is even we have $\frac{1}{2}\cdot(f(x) + f(-x) ) = f(x)$. 
Moreover, $\frac{1}{2}\cdot (p(x) + p(-x) )$ is an even polynomial already. We are done.
A: Just for reference, here is another trick: Since $x\mapsto f(\sqrt{x})$ is continuous on $[0, 1]$, for $\epsilon > 0$ there exists a polynomial $q(x)$ such that
$$ \sup_{x\in[0,1]} |f(\sqrt{x}) - q(x)| < \epsilon. \tag{*}$$
Now let $p(x) = q(x^2)$ and notice that $p$ is even and
$$ \sup_{x\in[-1,1]}|f(x) - p(x)| = \sup_{x\in[0,1]} |f(\sqrt{x}) - q(x)| \stackrel{\text{by (*)}}{<} \epsilon $$
A: A little late, but I'd like to add a short proof for possible future readers. The subspace $E$ of even polynomials on $[0,1]$ is dense in $C([0,1])$ by Stone-Weierstrass.  Approximate $f|_{[0,1]}$ by $p\in E$ to within $\epsilon$; then $x\mapsto p(|x|)$ on $[-1,1]$ is an even polynomial, and $|f(x)-p(|x|)|<\epsilon$ on $[-1,1]$.
