Degree of unsaturation of naphthalene
You determined the number of double bonds incorrectly, probably due to structure representation with aromatic rings, which, as Loong pointed out, is not entirely correct.
You can redraw the structure of naphthalene as such:



and make sure that there are five double bonds and two rings giving degree of unsaturation

$$\text{DU} = n_\text{rings} + n_\text{π-bonds} = 2 + 5 = 7$$

You can also compare formula of naphthalene $\ce{C10H8}$ with fully saturated alternative $\ce{C10H22}$ and proceed from there: 

$$\text{DU} = (H_\mathrm{sat} - H)/2 = (22 - 8)/2 = 7$$

You again arrive at $\text{DU} = 7$ using a simplified formula:

$$\text{DU} = C - H/2 + 1 = 10 - 8/2 + 1 = 7$$