Why is the polynomial best approximation to an even function itself even? I have seen this stated and it seems intuitively obvious but I cannot prove it. I have a feeling it may be because a non-even best approximant would not satisfy the equioscillation property of the best approximation (that if |f-p|=d, where p is the best polynomial approximation of order n, there are at least n+2 successive points in the interval under consideration where alternately |f-p|=+-d ).
Note: By "best approximation" I mean the polynomial that minimizes |f-p| on [-1,1] where |.| is the sup norm. So e.g. the best approximation of order n would be p in Pn={polynomials degree<=n} which does this.

First, we must pin down what we mean by "better approximate." If you are interested in the interval $[-1,1]$ and your given function is $f(x)$, one reasonable definition of approximation error by a second function $g(x)$ is
$$\int_{-1}^1 [f(x)-g(x)]^2\,dx.$$
Now suppose $f(x)$ is an even function, and $e(x)$ an even polynomial approximation to $f(x)$. Can we improve the approximation by adding some odd polynomial terms $o(x)$? Let's check:
\begin{align*}
\int_{-1}^1 [f(x)-e(x)-o(x)]^2\,dx &= \int_{-1}^1 [f(x)-e(x)]^2 - 2[f(x)-e(x)]o(x) + o(x)^2\,dx\\
&= \int_{-1}^1 [f(x)-e(x)]^2\,dx + \int_{-1}^1 o(x)^2\,dx -2\int_{-1}^1[f(x)-e(x)]o(x)\,dx.
\end{align*}
Now let's use the fact that $f$ and $e$ are even, and $o$ is odd:
$$\int_{-1}^1[f(x)-e(x)]o(x)\,dx =  -\int_0^1[f(x)-e(x)]o(x)\,dx + \int_0^1 [f(x)-e(x)]o(x)\,dx = 0.$$
Therefore 
$$\int_{-1}^1 [f(x)-e(x)-o(x)]^2\,dx = \int_{-1}^1 [f(x)-e(x)]^2 + \int_{-1}^1 o(x)^2\,dx \geq \int_{-1}^1 [f(x)-e(x)]^2$$
and you were better off without the odd terms.
EDIT: For different norms, you carry out different flavors of the same argument. For instance, for the $\sup$ norm, 
\begin{align*}
\sup_{x\in [-1,1]} |f(x)-e(x)-o(x)| &= \sup_{x\in [0,1]} \max\left(|f(x)-e(x)-o(x)|,|f(x)-e(x)+o(x)|\right)\\
&\geq \sup_{x\in [0,1]} |f(x)-e(x)|\\
&= \sup_{x\in [-1,1]} |f(x)-e(x)|.
\end{align*}
