The methyl group is not supposed to be attacked. Elimination reactions require strong bases that react by removng a hydrogen proton (the hydrogen bond cleaves heterolytically).
Even by showing the methyl substituent being removed, the problem is not correct in doing so even at that, because only the carbon next to the carbon bearing the OH group can be attacked. This is the only way to form our double bond.
But yes the product does not have as many carbons as you began with, and since elimination reactions do not involve removal of carbons, this is a clear way to see that this reaction is not going to work that way.