I'm reading Halmos's Finite Dimensional Vector Spaces, in which he makes several references to the infinite dimensional case. In my edition this item appears as question 6 at the end of section 14.
The proof for a finite dimensional space is fairly straightforward: y and z are linear functionals and span at most one dimensional subspaces of the dual space V', say Y and Z respectively. The annihilators Y$^0$ and Z$^0$ exist in the double dual V''. For a finite dimensional space V`` is "equal" (isomorphic) to V and the condition [x,z] = 0 $\implies$ [x,y] = 0, equates to Y$^0$ is a subspace of Z$^0$, and they are both either of the same dimension as V, or one less. This means that the vectors x (if any) for which [x,y] and [x,z] are non-zero exist in the same one dimensional subspace of V, so that if x$_0$ is a basis for this (1 dimensional) space and $\alpha $ = [x$_0$,y]/[x$_0$,z] then [x$_0$,z] $\alpha $ = [x$_0$,y], and if [w,y] is non-zero then w = $\beta$x$_0$ (because w is in the 1-D space with basis x$_0$) so that [w,y] = [$\beta$x$_0$,y] = $\beta$[x$_0$,y] = $\beta$[x$_0$,z] $\alpha $ = [$\beta$x$_0$,z] $\alpha $ = [w,z] $\alpha$
So, my question is, is this true in an infinite dimensional space ? I can't prove it or find a counter-example.