Let us call the weighted $L^2$ space $W$, assume all the polynomials belong to $W$ and let $\Pi_n$ denote the subspace of polynomials of degree $n$ or less.
The essential point to note is that any $p \in \Pi_n$ is the closest approximation for $f \in W$ (in the sense of the $W$ norm) if and only if $f-p$ is orthogonal $\Pi_n$. If $p_0, p_1, \cdots, p_n, \cdots$ are the $W$ orthogonal polynomials each with unit norm, then the $n$th order approximation
\begin{align*}
f_n = \sum_{k=0}^n \langle f, p_k \rangle p_k
\end{align*}
is optimal because $p_1, \cdots p_n$ spans $\Pi_n$ and $f-f_n$ is easily verified to be in $\Pi_n^\perp$. Nor is it hard to see that such an expansion is unique. In particular the $n$th degree Taylor series cannot be any better. Note it might be no worse either, an obvious case being when $f$ is itself a polynomial of degree less than or equal to $n$.
The key result is: if $H$ is an inner product space, $x \in H$ and $V$ a subspace of $H$, then $x \in V^\perp$ if and only if
$$ \lVert v-x \lVert \geqslant \lVert x \rVert$$ for every $v \in V$. To prove it, first, assume $x \in V^\perp$, then for any $v \in V$,
\begin{align*}
\lVert v-x\rVert^2 &= \langle v-x, v-x\rangle \\
&= \lVert v \rVert^2 + \lVert x \rVert^2 \\
&\geqslant \lVert x \rVert^2
\end{align*}
Conversely if $\lVert v -x \rVert \geqslant \lVert x \rVert$ for all $v \in V$ then for any $u \in V$ and $\alpha \in \mathbb C$, we also have $\alpha u \in V$ and
\begin{align*}
\lVert x \rVert ^2 &\leqslant \lVert x - \alpha u \rVert ^2 \\
&= \lVert x \rVert x ^2 - \alpha \langle u, x \rangle - \overline{\alpha \langle u, x \rangle} +|\alpha|^2 \lVert u \rVert^2 \\
&= \lVert x \rVert^2 - 2 \Re \Big( \alpha \langle u, x\rangle \Big) + |\alpha|^2 \lVert u \rVert ^2 \tag{1}\label{BPA-1}
\end{align*}
Now choose $\theta$ so that $e^{i\theta}\langle u, x \rangle $ is real and positive and for any $r > 0$ let $\alpha = re^{i\theta}$. Then cancel $\lVert x \rVert^2$ on each side of inequality \eqref{BPA-1}, then divide by $r > 0$, to get
\begin{align*}
2 \lvert \langle u, x \rangle \rvert \leqslant r \lVert u \rVert^2.
\end{align*}
Now $r$ is arbitrary, so we must have $\langle u, x \rangle = 0$ and $u \in V^\perp$.
