Suppose that $X$ is a Banach space and $P$ and $Q$ be bounded linear projections on $X$ such that $PQ$ and $QP$ are compact. Does it follow that $PQ$ and $QP$ are finite-rank operators?
My attempt: I claim that both $PQ$ and $QP$ have closed range so if the range of one of them were not finite-dimensional, we would find a bounded sequence in it without a convergent subsequence.
Is it fine? If so, can we find a projection $R$ with finite-dimensional range such that $PQ$ and $QP$ commute on the image of $I-R$?
Let $H$ be an infinite-dimensional separable Hilbert space. Let $\{E_{kj}\}$ be the canonical set of matrix units. Let $$ X=\sum_{k=1}^\infty \frac1k\,E_{2k,2k+1}, \ \ \ E=\sum_{k=1}^\infty E_{2k,2k} $$ and let $P=E+X$, $Q=I-E+X^*$. Note that $X^2=0$, $EX=X$, and $XE=0$, so $$ P^2=P, Q^2=Q. $$ Also, $$ PQ=E(I-E)+EX^*+X(I-E)+XX^*=X+XX^*, $$ $$ QP=(I-E)E+(I-E)X+X^*E+X^*X=X^*+X^*X. $$ So both $PQ$ and $QP$ are compact, but not finite-rank.
It may help to see them as direct sums: $$ P=0\oplus\bigoplus_{k=1}^\infty\begin{bmatrix}1&1/k\\0&0\end{bmatrix},\ \ Q=1\oplus\bigoplus_{k=1}^\infty\begin{bmatrix}0&0\\1/k&1\end{bmatrix}, $$ and then $$ PQ=0\oplus\bigoplus_{k=1}^\infty \begin{bmatrix}1/k^2&1/k\\0&0\end{bmatrix}, $$ $$ QP=0\oplus\bigoplus_{k=1}^\infty \begin{bmatrix}0&0\\1/k&1/k^2\end{bmatrix}. $$