Below I have drawn a molecule with random substituents $\ce{X}$ and $\ce{Y}$:
If the distances from $\ce{H_b}$ and $\ce{H_a}$ to $\ce{H_d}$ are considered, they are different - $\ce{H_b}$ is 5 bonds away and $\ce{H_a}$ is 3 bonds away, so $\ce{H_b}$ sees $\ce{H_d}$ differently compared to $\ce{H_c}$. The same argument can be made vice versa and also between the same pair of protons towards $\ce{H_c}$. This was what the lecturer told me regarding this compound below.
This led me to believe then that all the protons in the molecule are magnetically inequivalent and using up to $\ce{^4J}$ coupling, all protons show a doublet of doublet.
However, when I was in school, I was told that if symmetry was a property of a molecule, as above, $\ce{H_a}$ and $\ce{H_b}$ were equivalent, as are $\ce{H_c}$ and $\ce{H_d}$, and each pair of protons gives a doublet. Researching this matter on Google still held true in my undergraduate studies.
What am I getting wrong here?
I read a quote on another question highlighted here
For two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other non-chemical shift equivalent nuclei in the molecule.
So as $\ce{H_a}$ and $\ce{H_b}$ are chemically equivalent but $\ce{H_b}$ couples differently to $\ce{H_c}$ than $\ce{H_a}$ does, that therefore means that surely $\ce{H_a}$ and $\ce{H_b}$ are magnetically inequivalent?