 Solution

An arbitrary orthonormal basis does not have the printed property. For example on $[-1,1]$ with $w=1$, take the usual normalized constant and linear Legendre polynomials and swap their labels $p_0,p_1$. The resulting family is still an orthonormal basis but $p_0$ has one zero, rather than zero. The necessary intended hypothesis is that $p_k$ has degree $k$, or equivalently that the basis is obtained by orthogonalizing $1,x,x^2,\dots$ in that order.

Under this hypothesis, $p_k$ is orthogonal to every polynomial of degree less than $k$. Let $t_1,\dots ,t_s$ be its sign-changing zeros in $(a,b)$ and put $q(x)={\prod }_{j=1}^s(x-t_j)$. If $s<k$, then $p_{k}q$ has a constant nonzero sign except at finitely many points: each interior sign change has been canceled, and all remaining zeros have even multiplicity or occur at endpoints. Since $w>0$, its integral cannot vanish. But orthogonality to $q$, of degree $s<k$, says that integral is zero, a contradiction. Thus $s\ge k$. Since the polynomial has degree $k$, it follows thatThis proves the requested conclusion in its standard intended setting and explains why it does not follow from the literal basis statement alone.