= Solution
Let $A_n$ be finite-rank bounded operators with $\|T-A_n\|\to0$ in <operator norm>. Fix $\epsilon>0$ and choose $n$ with $\|T-A_n\|<\epsilon/3$. The set $A_n(B_X)$ is bounded in a finite-dimensional <vector subspace>[subspace], hence has a finite $\epsilon/3$-net $y_1,\ldots,y_m$.
For each $x\in B_X$, some $j$ satisfies
$$
\|Tx-y_j\|\leq\|(T-A_n)x\|+\|A_nx-y_j\|<2\epsilon/3.
$$
Thus $T(B_X)$ is <totally bounded>. Its <closure> is <totally bounded> and, as a closed subset of the <Banach space> $Y$, <complete>. A <complete> <totally bounded> metric space is <compact>. Consequently \b[$T$ is <compact>].
The operator-norm limit is bounded, since $\|T\|\leq\|A_n\|+\|T-A_n\|$ for any sufficiently good approximation. This proves directly that an <approximable operator> into a <Banach space> is <compact>; no assertion that every <compact operator> on an arbitrary <Banach space> is approximable is needed.
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