= Solution
For $x=\sum_nx_ne_n$, <Cauchy-Schwarz inequality> gives
$$
\|Tx\|
\leq\sum_n|x_n|\,\|Te_n\|
\leq\|x\|\left(\sum_n\|Te_n\|^2\right)^{1/2}.
$$
Thus every $T\in H(V,V)$ is bounded and
$\|T\|\leq\|T\|_*$. The truncations
$$
T_Nx=\sum_{n\leq N}x_nTe_n
$$
have finite rank and satisfy
$$
\|T-T_N\|\leq
\left(\sum_{n>N}\|Te_n\|^2\right)^{1/2}\to0.
$$
Hence every such <Hilbert-Schmidt operator> is compact.
Define
$$
\langle S,T\rangle_*=\sum_n\langle Se_n,Te_n\rangle.
$$
This is an inner product inducing $\|\cdot\|_*$. If $(T_j)$ is Cauchy in this norm, then each $T_je_n$ converges to some $y_n$, and the usual Fatou/tail argument gives
$\sum_n\|y_n\|^2<\infty$ and
$\sum_n\|T_je_n-y_n\|^2\to0$. Defining
$Tx=\sum_nx_ny_n$ produces a bounded operator by the preceding estimate and gives
$T_j\to T$ in $\|\cdot\|_*$. Thus $(H(V,V),\|\cdot\|_*)$ is a Hilbert space.
The norms are not equivalent in infinite dimension. The rank-$N$ orthogonal projection has
$$
\|P_N\|=1,\qquad \|P_N\|_*=\sqrt N.
$$
Although $\|T\|\leq\|T\|_*$, no uniform reverse inequality can hold.
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