= Solution
Construct compatible choices in the nested spaces, rather than choosing unrelated <orthonormal bases> for different $V_n$. Set $V_{-1}=\{0\}$ and define
$$
L_n=V_n\cap V_{n-1}^{\perp}.
$$
The <orthogonal complement> is taken for the <inner product> from part (i). Since $\dim V_n=n+1$ and $\dim V_{n-1}=n$, the space $L_n$ has dimension one and $V_n=V_{n-1}\oplus L_n$.
It is an <invariant subspace>. Indeed, if $v\in L_n$ and $u\in V_{n-1}$, part (ii) gives
$$
\langle u,Tv\rangle=\langle Tu,v\rangle=0,
$$
because $Tu\in V_{n-1}$. Also $Tv\in V_n$. Hence $Tv\in L_n$. Any nonzero vector in this one-dimensional space is consequently an <eigenvector>. Choose a unit vector $g_n\in L_n$. It has <polynomial degree> exactly $n$: membership in $V_{n-1}$ would make it orthogonal to itself and thus zero. Part (iii) now supplies its <eigenvalue>.
For $m<n$, $g_m\in V_{n-1}$ and $g_n\perp V_{n-1}$, so the choices are pairwise orthogonal. We have proved
$$
\boxed{\langle g_m,g_n\rangle=\delta_{mn},\qquad
\deg g_n=n,\qquad Tg_n=-2(n+1)g_n\quad(n\geq0).}
$$
An explicit compatible construction is the <Gram-Schmidt process> on $1,x,x^2,\ldots$:
$$
r_n=x^n-\sum_{j=0}^{n-1}\langle g_j,x^n\rangle g_j,
\qquad g_n=\frac{r_n}{\sqrt{\langle r_n,r_n\rangle}}.
$$
The leading term of $r_n$ is $x^n$, so it is never zero. This fixes the positive real leading coefficient and yields the normalized <Hermite polynomials>. For example, the <Gaussian integral> gives
$$
g_0=\pi^{-1/4},\qquad
g_1=\sqrt2\,\pi^{-1/4}x,\qquad
g_2=\frac{2x^2-1}{\sqrt2\,\pi^{1/4}}.
$$
Every <polynomial> is a finite linear combination of these <eigenvectors>, since $g_0,\ldots,g_n$ form an <orthonormal basis> of $V_n$ for each $n$.
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