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If $A$ has finite rank, the image of its unit ball is bounded in the finite-dimensional space $\mathrm{ran}A$. The closure of a bounded set in a finite-dimensional normed space is compact. Hence every bounded finite-rank operator is a compact operator.

Let $Q_n=P_n^{\ast }{P}_n$ be the orthogonal projection ontoThen $Q_n\to I$ strongly. Strong convergence is uniform on every compact subset: if $K$ is compact, cover it by finitely many small balls and use $\Vert I-Q_n\Vert \le 1$ at their centers. Since the closure of $A$ applied to the unit ball is compact,The adjoint of a compact operator is compact, so the same argument for $A^{\ast }$ givesSince $A_n=Q_{n}A{Q}_n$,ThereforeThis is the finite-section approximation of a compact operator.