= Solution
A <bounded linear operator> $T:X\to Y$ is a <compact operator> when the image of the closed unit ball has <compact closure>, equivalently when every bounded sequence $(x_n)$ has a subsequence for which $(Tx_n)$ converges.
Each $T_i$ is compact because its image lies in a finite-dimensional subspace. Let $B_X$ be the unit ball and fix $\varepsilon>0$. Choose $i$ with $\lVert T-T_i\rVert<\varepsilon/3$. Since $T_i(B_X)$ is relatively compact, it has a finite $\varepsilon/3$-net $y_1,\ldots,y_N$. For each $x\in B_X$, choose $y_j$ with $\lVert T_ix-y_j\rVert<\varepsilon/3$; then
$$
\lVert Tx-y_j\rVert
\leq\lVert(T-T_i)x\rVert+\lVert T_ix-y_j\rVert
<\frac{2\varepsilon}{3}.
$$
Thus $T(B_X)$ is <totally bounded>. Its closure is complete because $Y$ is a <Banach space>, so it is compact. This proves the <norm-closedness of compact operators> and hence
$$
\boxed{T\text{ is compact}.}
$$
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