Why is the ether linkage breaking here? I have been asked to find the major product of the following reaction:

This is my thought process: the lone pairs of the oxygen atom connected to $\ce{-Et}$ are less delocalised than the other oxygen atom. So, that will abstract the $\ce{H^+}$ from $\ce{HI}$ leaving behind $\ce{I^-}$ and producing a positive charge on the oxygen atom. Now the oxygen atom will take up the bond pair electrons and $\ce{I^-}$ will form a bond in $\ce{S_{N}2}$ mechanism, producing $\ce{I-Ph-O-Ph}$ and $\ce{EtOH}$ as products.

But, the answer in the book states that:

The ether linkage will break as it does not have a double bond character due to the absence of resonance. 

Why is the ether linkage breaking here? What does it mean by &quot;it does not have a double bond character&quot;?
I have used the image you used $2nd$, see carefully the markings.
The $C-O$ bond with benzene ring is stronger than the $C-O$ bond of ethyl, as there is resonance with benzene making it partially a double bond. So, that $C-O$ is difficult to break compared to the $C-O$ bond in ethyl branch.