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Use the standard convention that the orthogonal polynomials are nonzero. Their exact degrees then follow from the stated assumptions by induction: ${\varphi }_0$ is a nonzero constant, and if ${\varphi }_0,\dots ,{\varphi }_{r-1}$ span $P_{r-1}$, a nonzero ${\varphi }_r\in P_r$ orthogonal to all of them cannot lie in $P_{r-1}$.

Let $t_1,\dots ,t_m$ be the distinct sign-changing roots of ${\varphi }_r$ in $(0,1)$, and put $q(x)={\prod }_{j=1}^m(x-t_j)$. Then ${\varphi }_{r}q$ has constant sign between and across all its roots, and is not identically zero. Because $w>0$ in the interior, ${\int }_0^{1}w{\varphi }_{r}q$ is nonzero. If $m<r$, orthogonality to $P_{r-1}$ would make this integral zero. Thus $m\ge r$.

A degree-$r$ polynomial cannot have more than $r$ roots counted with multiplicity, so all $r$ roots are real, simple, and strictly inside $(0,1)$. This is the standard zeros of orthogonal polynomials argument, with the positive weight ensuring strict noncancellation.

Let $t_1,\dots ,t_m$ be the distinct roots in $(0,1)$ at which ${\varphi }_r$ changes sign, and put $q(x)={\prod }_{j=1}^m(x-t_j)$. Both ${\varphi }_r$ and $q$ change sign at each such point, while roots of even multiplicity do not change sign. Consequently ${\varphi }_{r}q$ has one constant sign on $(0,1)$ except at its zeros, and is not identically zero. Its integral cannot vanish.

If $m<r$, however, orthogonality of the monic orthogonal polynomial ${\varphi }_r$ to every polynomial of lower degree gives $({\varphi }_r,q)=0$, a contradiction. Hence $m\ge r$. Since $\mathrm{deg}{\varphi }_r=r$, there must be exactly $r$ such roots, each of multiplicity one. They exhaust the degree, leaving no complex, exterior, or endpoint roots. Thus all roots are simple and lie strictly in $(0,1)$, which proves the requested weaker inclusion in $[0,1]$. This is the sign-change proof of the zeros of orthogonal polynomials theorem.