Let $H$ be a Hilbert space and let $F$ be a closed subspace. Then I'm to prove that the projection $p:H\rightarrow F$ is a compact operator if and only if $F$ is finite-dimensional. It's easy to prove that if $F$ is finite-dimensional, then $p$ is compact; since $p$ is a bounded operator with finite rank the result follows immediately.
Now the proof in the other direction is a different story. To argue by contradiction, I should assume the projection operator is compact, but suppose that $F$ is infinite-dimensional for a contradiction. Now all I need to do is give a bounded sequence $(x_n)$ in $H$ such that $(px_n)$ has no convergent subsequence.
However I think I should be able to prove it directly. If I can show that $p$ compact implies $F$ locally compact, then since $F$ is a locally compact Hausdorff topological space it must be finite-dimensional, and I'm done. So, trying to prove this implication, I let $x \in F$ and try to show it has a compact neighborhood. Now, if I let $U$ be a bounded neighborhood around $x$ and let $(y_n)$ be any sequence in $U$, then $py_n = y_n$, so $(y_n)$ must have a convergent subsequence.
Is my sketch in the last paragraph correct or is that too easy?