= Solution
Use the intended space of <Hilbert-Schmidt operators>, with the given <Hilbert-Schmidt inner product> linear in its first argument. It remains to prove completeness in its <Hilbert-Schmidt norm>. Let $(T_n)$ be a <Cauchy sequence> in that <norm>. The bound $\|T_n-T_m\|\leq\|T_n-T_m\|_2$ makes it a <Cauchy sequence> in <operator norm>. The space $\mathcal B(X)$ is a <Banach space>, so $T_n\to T$ in <operator norm> for some <bounded linear operator> $T$.
For completeness, that last standard fact follows by taking $Tx=\lim_nT_nx$ for each $x$: the <Cauchy sequence> $(T_n)$ is uniformly bounded in <operator norm>, so its limit is linear and bounded, and the uniform Cauchy estimate on the <unit ball> gives <operator norm> convergence.
Given $\varepsilon>0$, choose $N$ so that $\|T_n-T_m\|_2<\varepsilon$ for $n,m\geq N$. Fix $n\geq N$ and any positive integer $K$. Since $T_me_k\to Te_k$ for each $k$, passage to the limit in a finite sum gives
$$
\sum_{k=1}^K\|(T_n-T)e_k\|^2
=\lim_{m\to\infty}\sum_{k=1}^K\|(T_n-T_m)e_k\|^2
\leq\varepsilon^2.
$$
Taking the supremum over $K$ gives $\sum_k\|(T_n-T)e_k\|^2\leq\varepsilon^2$. Also,
$$
\sum_k\|Te_k\|^2
\leq2\sum_k\|T_Ne_k\|^2+2\sum_k\|(T-T_N)e_k\|^2<\infty.
$$
Thus $T\in S_2(X)$, and $\|T_n-T\|_2\leq\varepsilon$ for $n\geq N$. This proves <Hilbert-Schmidt operators form a Hilbert space>:
$$
\boxed{S_2(X)\text{ is complete in }\|\cdot\|_2,\text{ hence is a Hilbert space}.}
$$
Under the literal unrestricted “linear mappings” definition, the nonzero $R$ from part (i) satisfies $\langle R,R\rangle_{S_2(X)}=0$. The form would fail positive definiteness and would not be an <inner product>; the boundedness correction is needed here too.
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