= Solution
A real <inner product> is a bilinear symmetric form satisfying $(v,v)>0$ for every nonzero $v$, and $(0,0)=0$. Its <norm> is $\|v\|=\sqrt{(v,v)}$. An <orthogonal polynomial> family has nonzero members of exact degrees $0,1,\ldots,k$, with $(\phi_i,\phi_j)=0$ for different indices. Exact degree is part of this definition; merely belonging to the indicated <polynomial vector spaces> would not suffice.
Distinct degrees make $\phi_0,\ldots,\phi_{\ell-1}$ <linearly independent>. There are $\ell$ of them in the $\ell$-dimensional space $\Pi_{\ell-1}$, so they form a <basis>. Any lower-degree <polynomial> is their <linear combination>, and bilinearity therefore gives $(\phi_\ell,p)=0$ for every $p\in\Pi_{\ell-1}$.
Use the orthogonal-projection characterization: $p_*$ is the <best approximation> to $v$ in a finite-dimensional subspace $W$ exactly when $v-p_*$ is <orthogonal> to $W$. Here $v=\phi_\ell$ is already <orthogonal> to $W=\Pi_{\ell-1}$, so its unique <best approximation> is zero. For $\ell\ge1$, $\phi_{\ell-1}$ is nonzero and therefore cannot be that <best approximation>. Directly,
$$
\|\phi_\ell-\phi_{\ell-1}\|^2
=\|\phi_\ell\|^2+\|\phi_{\ell-1}\|^2>\|\phi_\ell\|^2.
$$
For the specified weight, symmetry makes every odd moment vanish, while
$$
\int_{-1}^1|x|x^{2m}\,dx=2\int_0^1x^{2m+1}\,dx=\frac1{m+1}.
$$
Choose $\phi_0=1$ and $\phi_1=x$. A monic quadratic <orthogonal> to $x$ has no linear term by symmetry, and <orthogonality> to 1 requires $\int|x|(x^2-c)dx=1/2-c=0$. Thus a valid answer is
$$
\boxed{\phi_0=1,\qquad\phi_1=x,\qquad\phi_2=x^2-\tfrac12.}
$$
The weight vanishes only at one point, so a nonzero <polynomial> has positive squared <norm>; the form is indeed positive definite on the <polynomial vector space>. Nonzero rescalings of individual family members give equally valid unnormalized answers.
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