What is the oxidation state of copper in Gilman's Reagent?
Don't count bonds.  Count electrons.

Copper has $11$ valence electrons as a neutral atom, as the $d$ orbitals are able to participate in reactions.  In the oxidation state formalism the alkyl groups, whose atoms are more electronegative than copper, are each considered to take an electron away from the copper to complete their carbon octets.  But then the lithium, bonded to copper (bonding lithium to the alkyl groups would give a separate organolithium molecule instead), adds an electron.  So the copper goes from $11$ valence electrons down to $10$ in the oxidation state formalism, losing one overall; the oxidation state of copper is thereby $+1$.