If you insist on neutral atoms then the internal linkage is not conjugated.
BUT ...
The notable thing about azulene is that the seven-and five-membered rings can be aromatic when the former is positively charged and the latter negatively charged. That puts six electrons into each ring, and we all know about the $4n+2$ rule (even if technically it should be applied to single-ring systems).
If you allow a positive charge on the seven-membered ring and a negative charge on the five-membered ring, you can render contributing structures with a double bond in the internal linkage.
![enter image description here](https://i.sstatic.net/00TeUm.jpg)
http://www.chemspider.com/Chemical-Structure.8876.html
According to Wikipedia, azulene is actually known with a dipole moment having the positive end in the seven-membered ring, and its chemical reactivity (electrophilic in the seven-membered ring, nucleophilic in the five-membered ring) is also consistent with this polar contributing structure.
C(C=CC(=CC=CC)C(C=CC=C)=CC=CC)=C
So, the "internal" bond of azulene might not be conjugated, but both carbon atoms that are part of that internal bond are conjugated for other reasons... $\endgroup$