= Solution
Let $(T_m)$ be a <Cauchy sequence> in the <operator norm>. For each $v\in V$,
$$
\|T_mv-T_nv\|\leq\|T_m-T_n\|\,\|v\|,
$$
so $(T_mv)$ is a Cauchy sequence. <Completeness> of the <Banach space> $V$ permits the definition
$Tv=\lim_mT_mv$. Pointwise passage to the <limit> proves that $T$ is a <linear map>[linear]. A norm-Cauchy sequence is a <bounded sequence>, say $\|T_m\|\leq C$, and therefore
$\|Tv\|\leq C\|v\|$, so $T$ is bounded. Finally, letting $m\to\infty$ in
$\|(T_n-T_m)v\|\leq\varepsilon\|v\|$ gives
$\|T_n-T\|\leq\varepsilon$. Thus $T_n\to T$ in operator norm and
$\mathcal B(V,V)$ is Banach.
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