 Solution

Take $S=T(B_X)$. A bounded linear operator is compact exactly when the closure of this set is compact. In a complete metric space this is equivalent to its being totally bounded.

If $T$ is compact, choose a finite $\epsilon$-net $z_1,\dots ,z_m$ for $T(B_X)$ and let $Y$ be their span. Then $\mathrm{dist}(Tx,Y)<\epsilon$ for $\Vert x\Vert \le 1$, giving $\Vert {\pi }_{Y}T\Vert \le \epsilon$.

Conversely, fix $\delta >0$ and take $Y$ with $\Vert {\pi }_{Y}T\Vert \le \delta /4$. For every $x\in B_X$ choose $y_x\in Y$ with $\Vert Tx-y_x\Vert <\delta /2$. These approximants satisfy $\Vert y_x\Vert \le \Vert T\Vert +\delta /2$, so form a bounded subset of a finite-dimensional space and have a finite $\delta /2$-net. The same centers give a finite $\delta$-net for $T(B_X)$. Thus it is totally bounded and $T$ is compact. In particular